Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T = [tex]\sqrt[4]{P/Ae}[/tex]
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
Which of these charges is experiencing the electric field with the largest magnitude? A 2C charge acted on by a 4 N electric force. A 3C charge acted on by a 5N electric force. A 4C charge acted on by a 6N electric force. A 2C charge acted on by a 6N electric force. A 3C charge acted on by a 3N electric force. A 4C charge acted on by a 2N electric force. All of the above are experiencing electric fields with the same magnitude
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) A 2C charge acted on by a 4 N electric force
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) A 3 C charge acted on by a 5 N electric force
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) A 4 C charge acted on by a 6 N electric force
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) A 2 C charge acted on by a 6 N electric force
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) A 3 C charge acted on by a 3 N electric force
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) A 4 C charge acted on by a 2 N electric force
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
A 2 C charge acted on by a 6 N electric force
HELP, END OF SCHOOL YEAR, 30 POINTS Unit 9 lesson 15 astronomy unit test answers I ONLY HAVE ONE MORE DAY
1 As evidence supporting the Big Bang theory, what does the redshift of light from galaxies indicate?
The universe is mainly hydrogen.
The universe is 13.8 billion years old.
The universe is cooling off.
The universe is expanding.
2 Which evidence supports the idea that Cosmic Microwave Background radiation is a remnant of the Big Bang?(1 point)
Its temperature is uniform.
Its mass fluctuates greatly.
Its temperature fluctuates greatly.
Its mass is uniform.
3 Which of these items provide evidence supporting the Big Bang theory? Select the two correct items.(1 point)
rate of star formation
composition of matter in the universe
sizes and shapes of distant galaxies
cosmic background radiation
4 How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory?(1 point)
The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.
The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.
The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.
The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.
5 How does weak background radiation coming from every direction in the sky support the Big Bang Theory?(1 point)
It provides evidence of the universe's increasing mass.
It provides evidence of universe expansion.
It provides evidence of universe contraction.
It provides evidence of the universe's decreasing mass.
6 Which statements describe ways that nuclear fission is different than nuclear fusion? Select the two correct answers.(1 point)
Nuclear fission is used to produce electricity at nuclear power plants.
Nuclear fission involves one large atom splitting into two smaller atoms.
Nuclear fission takes place in the nucleus of an atom.
Nuclear fission releases a huge amount of energy.
7 Blueshift is observed when(1 point)
a distant luminous object travels rapidly away from an observer.
a distant luminous object travels rapidly towards an observer.
a luminous object travels alongside an observer.
a luminous object is stationary compared to an observer.
8 Which statements about nuclear fusion are false? Select the two correct answers.(1 point)
The fuel for nuclear fusion is often uranium.
Nuclear fusion is used to generate electricity at nuclear power plants.
Nuclear fusion releases large amounts of energy.
Nuclear fusion takes place in the cores of stars.
9 Which of the following statements provide evidence to support the big bang theory? Select the two correct answers.
The ratios of hydrogen and helium in the universe match those of the early universe.
The universe began as a very high density singularity.
Dark matter makes up the majority of matter in the galaxy.
Small spiral galaxies become larger elliptical galaxies when they collide.
10 Which represents a correct match between ideas related to the formation of the universe? Select the two correct answers.(1 point)
accelerating expansion — dark energy
structures forming in the early universe — dark matter
greatest percent of mass of universe — dark matter
glowing nebulae — dark energy
11 How is dark energy related to the theory of the Big Bang?(1 point)
It causes the expansion of the universe to accelerate.
It causes the universe to expand.
It seeded the formation of galaxies and star clusters.
It causes the spinning of galaxies.
Answer:
1. The Universe is Expanding
2. It’s temperature is it’s uniform
3. Cosmic background radiation
4. I will give a hint for this one, since I don’t know, the hint is the universe is cooling.
5. It provides evidence of universe expansion.
6. Sorry I don’t know the rest
Explanation:
The universe is the collection of every item in space and time as well as the contents of those items
The correct options are as follows;
1. The universe is expanding
2. Its temperature is uniform
3. Composition of matter in the universe, cosmic background radiation
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
5. It provides evidence of universe expansion
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
7. A distant object travels rapidly towards an observer
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
9. The universe began as a very high density singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Structure forming in the early universe — Dark matter
11. It causes the expansion of the universe to accelerate
The reasons for selecting the above options are as follows;
1. The universe is expanding
The redshift of light from galaxies indicates that that are moving away
2. Its temperature is uniform
The uniform temperature of the microwave background suggest a common source
3. Composition of matter in the universe, cosmic background radiation
The matter present in the universe are characteristically similar in their origins
The cosmic background provides evidence of the existence of a singularity
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
Based on the Big Bang Theory, the temperature of the universe is reducing as the universe expands, compared to the initial temperature
5. It provides evidence of universe expansion
The background radiation coming from a single source as the rest of the universe is expected to spread throughout the universe
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
Nuclear fusion involves the joining of small atoms to form a larger atom
7. A distant object travels rapidly towards an observer
The redshift is the opposite, indicating that the object is moving further away
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
Nuclear fusion usually consists of joining small atoms together. It has not been used for commercial energy production
9. The universe began as a very high density singularity
According to the Big Bang Theory, the universe started from the dense, high temperature singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Dark energy causes expansion
Structure forming in the early universe — Dark matter
Dark matter is instrumental to the formation of structures in the universe
11. It causes the expansion of the universe to accelerate
Dark energy is seen as the cause of the accelerating expansion of the universe
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Describe how fractional distillation and cracking are used so that sufficient petrol is produced from crude oil to meet demand.
Answer:
☟
Explanation:
Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.
Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:
heated to 600-700°C
passed over a catalyst of silica or alumina
These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:
hexane → butane + ethene
C6H14 → C4H10 + C2H4
Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.
Fractional distillation of crude oil
Fractional distillation separates a mixture into a number of different parts, called fractions.
A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.
Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.
Oil fractions
The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.
As you go up the fractionating column, the hydrocarbons have:
lower boiling points
lower viscosity (they flow more easily)
higher flammability (they ignite more easily).
Other fossil fuels
Crude oil is not the only fossil fuel.
Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.
Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.
In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.
What is fractional distillation?Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.
Generally, the components have boiling points that differ by less than 25 °C from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.
The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.
Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.
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b) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected
in series in order that the same current shall be supplied from 240 V, 50 Hz mains.
Ignore the resistance of the inductor and calculate:
i. the inductance of the inductor;
ii. the impedance of the circuit;
iii. the phase difference between the current and the applied voltage.
Assume the waveform to be sinusoidal.
Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
For the RC circuit and the RL circuit, assume that the period of the source square wave is much larger than the time constant for each. Make a sketch of vR(t) as a function of t for each of the circuits?
Answer with Explanation:
Concepts and reason
The concept to solve this problem is that if a capacitor is connected in a RC circuit then it allows the flow of charge through circuit only till it gets fully charged. Once the capacitor is charged it will not allow any charge or current to flow.
Opposite is the case with inductor in the RL circuit. According to Faraday's law an inductor develops an emf to oppose the voltage applied but once the flux change stops then the inductor behaves just like a normal wire as if no inductor is there.
In attached figure, resistor is connected in series to the capacitor.
As we considered [tex]V_{C}[/tex] the voltage across the capacitor and [tex]V_{s}[/tex] the voltage across the source.
Voltage across a resistor In RC circuit.
[tex]V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )[/tex]
Voltage across a resistor In RL circuit.
[tex]V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )[/tex]
The sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits can be seen in the diagram attached below.
For the Pre-Laboratory exercise, based on the assumption that the RC circuit has a capacitor and a sensing resistor while the RL circuit has a sensing resistor and an inductor.
The input voltage for both circuits is regarded as the square wave and if the square wave is much larger than the time constant for each.
Therefore, we can conclude that the below diagram shows an appropriate sketch of [tex]\mathbf{v_R(t)}[/tex] as a function of t for each of the circuits.
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A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
Answer:
Explanation:
1. [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)
2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s
3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)
[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m
4. [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)
5. -
answer 4 could be wrong, not certain about that one and i don't know 5
"A trooper is moving due south along the freeway at a speed of 28 m/s. At time t = 0, a red car passes the trooper. The red car moves with constant velocity of 40 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.9 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car?"
Answer:
24.83 m
Explanation:
Applying the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = distance
v = velocity
t = time
a = acceleration
For the trooper;
v = 28 m/s
a = 2.9 m/s^2
Substituting into equation 1;
d1 = 28t + 0.5(2.9t^2)
d1 = 28t + 1.45t^2
For the red car;
v = 40 m/s
a = 0
Substituting into equation 1
d2 = 40t
The difference in distance is;
d = d2 - d1
d = 40t - (28t + 1.45t^2)
d = 12t - 1.45t^2
The maximum distance is at d(d)/dt = 0
differentiating d;
d' = 12 - 2.9t = 0
2.9t = 12
t = 12/2.9 = 4.137931034482
t = 4.138 s
Substituting t into function d;
d(max) = 12(4.138) - 1.45(4.138^2)
d(max) = 24.8275862 = 24.83 m
the maximum distance ahead of the trooper that is reached by the red car is 24.83 m
Un levantador de pesas puede generar 3000 N de fuerza ¿Cuál es el peso máximo que puede levantar con una palanca que tiene un brazo de la fuerza de 2 m y un brazo de resistencia de 50 cm?
Responder: 12000N
Explicación: Usando la fórmula para encontrar la eficiencia de una máquina. Eficiencia = ventaja mecánica / relación de velocidad × 100%
Dado MA = Carga / Esfuerzo
Relación de velocidad = distancia recorrida por esfuerzo (brazo de fuerza) / distancia recorrida por carga (brazo de resistencia)
MA = Carga / 3000
VR = 2 / 0.5 VR = 4
Asumiendo que la eficiencia es 100% 100% = (Carga / 3000) / 4 × 100%
1 = (Carga / 3000) / 4
4 = Carga / 3000
Carga = 4 × 3000
Carga = 12000N
Esto significa que el peso máximo que se puede levantar es 12000N
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to: A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:__________.
a) 19 s
b) 17 s
c) 21 s
d) 23 s
e) 15 s
Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is
[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]
For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],
[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]
which makes B, approximately 17 s, the correct answer.
The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)
Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:
[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.
[tex]\omega[/tex] - Final angular velocity, in radians per second.
[tex]\alpha[/tex] - Angular acceleration, in radians per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:
[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 16.667\,s[/tex]
The time interval of angular deceleration is 16.667 seconds. (Answer: B)
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A parallel-plate capacitor in air has a plate separation of 1.30 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and dis-connected from the source. The capacitor is then immersed in distilled water. Determine a) the charge on the plates before and after immersion.b) the capacitance and potential difference after immersion.c) the change in energy of the capacitor.
Answer:
Explanation:
capacitance of air capacitor
C = ε₀ A / d
ε₀ is permittivity of medium , A is plate area , d is distance between plate .
C = 8.85 x 10⁻¹² x 25 x 10⁻⁴ / 1.3 x 10⁻²
= 170.19 x 10⁻¹⁴ F .
charge on the capacitor when it is charged to potential of 255 V
= CV , C is capacitance and V is potential
= 170.19 x 10⁻¹⁴ x 255
= 4.34 x 10⁻¹⁰ C .
After it is disconnected from the source , and it is immersed in water , charge on it remains the same .
So its charge when immersed in water will be constant at 4.34 x 10⁻¹⁰ C.
b )
When it is immersed in water its capacity increases k times where k is dielectric constant of water which is 80 .
capacitance of capacitor in water = 80 x 170.19 x 10⁻¹⁴ F
= 13615.2 x 10⁻¹⁴ F .
= 1.36 x 10⁻¹⁰ F
potential difference = charge / capacitance
= 4.34 x 10⁻¹⁰ / 1.36 x 10⁻¹⁰
= 3.2 V
c )
Energy of capacitor = 1/2 C V²
Initial energy = 1/2 x 170.19 x 255² x 10⁻¹⁴
= 55.33 x 10⁻⁹ J
Final energy = 1/2 x 1.36 x 10⁻¹⁰ x 3.2²
= .7 x 10⁻⁹ J .
decrease of energy = 54.63 x 10⁻⁹ J .
A 550 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s
Answer:
[tex]52.25\times10^4W\\699.1 hp[/tex]
Explanation:
According to the energy conversation:
ΔK=[tex]-f_kd+W[/tex]
ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]
where,
[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.
[tex]v_i[/tex]= initial velocity of the system
[tex]v_f[/tex]=final velocity of the system
W= total work done on the system
[tex]f_k[/tex]= friction force
d= distance traveled
Given: [tex]v_f[/tex]=110m/s
d=400m
[tex]f_k[/tex]=1200N
[tex]v_i[/tex]=0m/s
t=7.3s
ΔK=[tex]-f_kd+W[/tex]
W= ΔK + [tex]f_kd[/tex]
=[tex]K_f-K_i+f_kd\\[/tex]
[tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]
[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]
Michelson and Morley's experiment is widely considered to have been:______
a. a success because it detected a shift in the interference pattern.
b. a failure because it detected a shift in the interference pattern.
c. a success because it did not detect a shift in the interference pattern.
d. a failure because it did not detect a shift in the interference pattern.
e. lacking the necessary precision to determine a shift in the interference pattern.
Answer:
The correct answer is option (c) a success because it did not detect a shift in the interference pattern.
Explanation:
In Michelson and Morley experiment it was considered to be successful.
They both found out that the experiment that was carried out was not a failure since it did not detect any shift in the interference pattern.
With this findings it was widely regarded as correct and precise.
A 5.50-kg bowling ball moving at 9.00 m/s collides with a 0.850-kg bowling pin, which is scattered at an angle of 85.0 0 to the initial direction of the bowling ball and with a speed of 15.0 m/s. (a) Calculate the final velocity (magnitude and direction) of the bowling ball.
Answer:
9.05 m/s , -14.72° (respect to x axis)
Explanation:
To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:
[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]
[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]
m1: mass of the bowling ball = 5.50 kg
m2: mass of the bowling pin = 0.850 kg
v1xi: initial velocity of the bowling ball = 9.0 m/s
v2xi: initial velocity of bowling pin = 0m/s
v1: final velocity of bowling ball = ?
v2: final velocity of bowling pin = 15.0 m/s
θ: angle of the scattered bowling pin = ?
Φ: angle of the scattered bowling ball = 85.0°
Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.
First you solve for v1cosθ in the equation for the x component of the momentum:
[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]
and also you solve for v1sinθ in the equation for the y component of the momentum:
[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]
Next, you divide v1cosθ and v1sinθ:
[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]
the direction of the bawling ball is -14.72° respect to the x axis
The final velocity of the bawling ball is:
[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]
hence, the final velocity of the bawling ball is 9.05 m/s
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
What is the speed at which a spaceship shoots up from earth ?
Answer:
Once at a steady cruising speed of about 16,150mph (26,000kph
Explanation:
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________
a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above
Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:
[tex]\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m[/tex]
The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
[tex]d=\sqrt{(x-x_o)^2+(y-y_o)^2}[/tex] (1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):
[tex]d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m[/tex]
hence, the displacement of the airplane is 3.45*10^4 m
An industrial flywheel (a solid disk) of mass 10.0 kg and radius 17.3 cm is rotating at an angular speed of 22.0 rad/s. Upon being switched to a slower setting, the flywheel uniformly slows down to 13.5 rad/s after rotating through an angle of 13.8 radians. Calculate the angular acceleration of the flywheel in the process of slowing down
Answer:
Explanation:
During slowing down , initial angular velocity ω₁ = 22 rad /s
final angular velocity ω₂ = 13.5 rad /s
using the law's of motion formula for rotation
ω₂² = ω₁² + 2 αθ , α is angular acceleration and θ is angle in radian rotated during this period
13.5² = 22² - 2xα x 13.8
2xα x 13.8 = 484 - 182.25
α = 10.93 rad / s²
What is The mass of an electron
What’s the answer to this question?
Answer:
6 A
Explanation:
Parallel connected resistors needs to be calculated as one single resistor. To do that: [tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]=[tex]\frac{3}{15}[/tex]=[tex]R^{-1}[/tex]
[tex]\frac{3}{15} ^{-1}[/tex]= 5 Ω (total resistance)
U = R* I
[tex]\frac{U}{R}[/tex]=I
[tex]\frac{30}{5}[/tex]=6 A
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. A particle with charge q is placed at the center of the cavity. The net charge on the inner surface of the conducting shell is
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
Although these quantities vary from one type of cell to another, a cell can be 2.2 micrometers in diameter with a cell wall 40 nm thick. If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?
Answer:
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
Explanation:
First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:
A = 4πr²
where,
A = Surface Area = ?
r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m
Therefore,
A = 4π(1.1 x 10⁻⁶ m)²
A = 15.2 x 10⁻¹² m²
Now, we find the volume of the cell wall. For that purpose, we use formula:
V = At
where,
V = Volume of the Cell Wall = ?
t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m
Therefore,
V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)
V = 60.82 x 10⁻²⁰ m³
Now, to find mass of cell wall, we use formula:
ρ = m/V
m = ρV
where,
ρ = density of water = 1000 kg/m³
m = Mass of Wall = ?
Therefore,
m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where
A = surface area of cell = 4πR² where R = radius of cell = 2.2 μm/2 = 1.1 × 10⁻⁶ m and t = thickness of cell wall = 40 nm = 40 × 10⁻⁹ m.Volume of cell wallSo, V = 4πR²t
Substituting the values of the variables into the equation, we have
V = 4πR²t
V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.
V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.
V = 193.6π × 10⁻²¹ m³
V = 608.21 × 10⁻²¹ m³
V = 6.0821 × 10⁻¹⁹ m³
V ≅ 6.082 × 10⁻¹⁹ m³
Mass of the cell wallWe know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.
Making m subject of the formula, we have
m = ρV
Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³
So, m = ρV
m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³
m = 6.082 × 10⁻¹⁶ kg
Converting to mg, we have
m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg
m = 6.082 × 10⁻¹⁰ mg
So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Learn more about mass of cell wall here:
https://brainly.com/question/13173768
A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simultaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?
If a metal rod is moved through magnetic field, the charged particles will feel a force, and if there is a complete circuit, a current will flow. We talk about the induced emf of the rod. The rod essentially acts like a battery, and the induced emf is the voltage of the battery. A magnetic field with a strength of 0.732 T is pointing into the page and a metal rod L=0.362 m in length is moved to the right at a speed v of 15.1m/s.
Required:
a. What is the induced emf in the rod?
b. Suppose the rod is sliding on conducting rails, and a complete circuit is formed. If the load resistance is 5.74Ω , what is the magnitude and direction (clockwise or counterclockwise) of the current flowing in the circuit?
Answer:
a. 4 V
b. 0.697 A
Explanation:
Magnetic field strength B = 0.732 T
length of rod l = 0.362 m
velocity of rod v = 15.1 m/s
a. EMF can be calculated as
E = Blv = 0.732 x 0.362 x 15.1 = 4 V
b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω
current I = V/R = 4/5.74 = 0.697 A
the current flows in a clockwise direction
A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.200 rad/s2 , what is its angular velocity at t = 2.50 s ? (Assume the acceleration and velocity have the same direction) Express your answer in radians per second. ω = nothing rads Request Answer Part B Through what angle has the wheel turned between t = 0 and t = 2.50 s ? Express your answer in radians. Δθ = nothing rad Request Answer Provide Feedback
Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.
(A) The wheel has angular velocity at time [tex]t[/tex] according to
[tex]\omega=\omega_0+\alpha t[/tex]
so that after 2.50 s, the wheel will have attained an angular velocity of
[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]
(B) The angular displacement of the wheel is given by
[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]
After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to
[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]
Two very large parallel sheets a distance d apart have their centers directly opposite each other. The sheets carry equal but opposite uniform surface charge densities. A point charge that is placed near the middle of the sheets a distance d/2 from each of them feels an electrical force F due to the sheets. If this charge is now moved closer to one of the sheets so that it is a distance d/4 from that sheet, what force will feel
Answer:
the force we will feel is F
Explanation:
According to the Gauss law, electric field due to very large sheet of charge is as follows.
[tex]E = \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
where,
[tex]\sigma[/tex] = charge per unit area
Since, it is given that there are two sheets of equal and opposite charge. Therefore, electric field between the plates will be as follows. [tex]E = \frac{\sigma}{2 \times \epsilon_{o}} + \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
Also, we know that relation between force and electric field is as follows.
F = qE
Hence, force felt by the charge present inside the plates will be as follows.
[tex]F = q \times \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
This depicts that force is not dependent on the distance and the charge is kept from one of the plate. Therefore, force F felt by the charge is same when it is placed at a distance d/2 and at a distance d/4 from one of the plate.
Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.
Answer:5.35nm
Explanation:
Consider that 1 inch is = 0.0254m
we have,
1m= 1x10^9 nm
While:
0.0254m = 2.54x10^7nm
1/55 (2.54x10^7) = 4.6181 x 10^5nm
1 day= 24 hrs
= (24x60) when calculating in min
= (24x60x60) calculating in seconds we have:
= 8.64x10⁴sec
In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm
Therefore, the amount by which the hair grows in 1 second will be;
= (4.6181 x 10^5)/(8.64x10^4)
= 5.35nm
The rate of growth will be 5.35nm