Answer:
663N
Explanation:
We need to find the force that will overcome the frictional force.
The angle of the normal force is 15°.
The mass of the crate is 32 kg
The coefficient of static friction is 0.49
Frictional force is given in terms of Normal force as:
F = μNcosθ
where μ = coefficient of static friction
N = normal force
θ = angle of normal force
Frictional force is given as:
F = mg
=>mg = μNcosθ
=> N = mg/(μcosθ)
N = (32 * 9.8) / (0.49 * cos15)
N= 313.6 / 0.473
N = 663 N
The force needed to cause the box to move must be 663N or greater.
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]
Answer:
Explanation:
The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]
What is equivalent resistance?The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.
What is equivalent resistance in series?Resistors are in series whenever the current flows through the resistors sequentially. It is given by
[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]
What is equivalent resistance in parallel?Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.
The equivalent resistance is the total resistance measured in a parallel. It is given by
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]
Given:
Resistor, [tex]R_{1} = 100 ohm[/tex]
Resistor, [tex]R_{2} = 200 ohm[/tex]
Voltage, [tex]V = 10 Volt[/tex]
Since, resistors are connected in parallel, the equivalent resistor is given by,
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]
[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]
[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]
Hence, the equivalent resistor is [tex]66.66 ohm[/tex].
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A balloon with a radius of 16 cm has an electric charge of 4.25 10 –9 C.
Determine the electric field strength at a distance of 40.0 cm from the balloon’s centre.
Answer:
239 N/C
Explanation:
Electric field strength at distance R from a charge Q is given by the expression
E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .
R = 40 cm , Q = 4.25 x 10⁻⁹
Putting the given values
E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²
= 239 N/C .
Which statement best describes one way that the molecules differ from atoms? a. A molecule can contain a nucleus about which its electrons orbit b. A molecule can contain two atoms of the same element. C. Only a molecule can be the smallest particle of a certain element. d. Only a molecule can be broken down into two or more different elements.
B and D are both true statements. I'm not comfortable saying that either one is better than the other one.
The statement that best describes one way that molecules differ from atoms is a molecule can contain two atoms of the same element, and only a molecule can be broken down into two or more different elements. The correct options are b and d.
What are atoms and molecules?According to science, an atom is the smallest component of an element that can exist freely or not. A molecule, on the other hand, is the smallest component of a chemical and is made up of a group of atoms linked together by a bond.
A molecule is the smallest component of a substance that has the chemical properties of the compound.
The term "independent molecule" is not commonly used to refer to atoms and complexes linked by non-covalent interactions such as hydrogen or ionic bonds. Molecules are common constituents of matter.
Therefore, the correct options are b and d.
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A swimmer heading directly through a 200m wide river reaches the opposite shore in 6 min 40s. She is washed downstream 480 m. How fast can you swim in calm water?
Answer :v=480m400s=1.2ms
2002+4802=H2
The hypotenuse H=520m
A quicker way to get the length of the hypotenuse is to recognize that this is a simple 5–12–13 triangle where the sides are multiples of 5, 12, and 13:
5(40) = 200m, 12(40)= 480m, 13(40)= 520m
We know that the swimmer travelled 520 m in 400 seconds, so her average speed was:
VR=520m400sec= 1.3ms
hope i got it right!! xx
Explanation:
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Answer:
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
Attractive
Explanation:
Data provided in the question
The potential energy of a pair of hydrogen atoms given by [tex]\frac{C_6}{X_6}[/tex]
Based on the given information, the force that one atom exerts on the other is
Potential energy μ = [tex]\frac{C_6}{X_6}[/tex]
Force exerted by one atom upon another
[tex]F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})[/tex]
or
[tex]F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})[/tex]
or
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
As we can see that the [tex]C_6[/tex] comes in positive and constant which represents that the force is negative that means the force is attractive in nature
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.
Answer:
A. xmax = 131.49 m
B. t = 8.74 s
C. ymax = 220.33 m
Explanation:
A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:
[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex] (1)
yo: height from the projectile is fired = 200m
vo: initial velocity of the projectile = 25m/s
g: gravitational acceleration = 9.8 m/s^2
θ: angle between the direction of the initial motion of the ball and the horizontal = 53°
t: time
You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.
When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:
[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)
You use the quadratic formula to obtain the value of t:
[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]
You use the positive value because it has physical meaning.
Now, you can calculate the horizontal range of the projectile by using the following formula:
[tex]x_{max}=v_ocos\theta t[/tex]
[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]
The cannon ball travels a horizontal distance of 131.49 m
B. The cannon ball reaches the canon for t = 8.74s
C. The maximum height is obtained by using the following formula:
[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex] (3)
By replacing in the equation (3) the values of all parameters you obtain:
[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]
The maximum height reached by the cannon ball is 220.33m
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.
The given parameters;
diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³The volume of the bobber is calculated as follows;
[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]
The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;
[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]
The volume of the lead is calculated as follows;
[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]
The buoyant force experienced by the lead due to the volume submerged is calculated as follows
[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]
The total buoyant force is calculated as;
[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]
The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;
[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]
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A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.
Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?
Answer:
a. i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Power generated is 3.6 W.
Explanation:
Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.
i.e V = IR
i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Given that; V = 12 V and R = 40 Ohm's.
P = IV
From Ohm's law, I = [tex]\frac{V}{R}[/tex]
So that;
P = [tex]\frac{V^{2} }{R}[/tex]
= [tex]\frac{12^{2} }{40}[/tex]
= [tex]\frac{144}{40}[/tex]
= 3.6 W
The power is 3.6 W.
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units
Answer:
Final Temperature = 71 °C
Explanation:
In this case, the ideal gas equation is written as;
PV = mRT
Where;
P is pressure
V is volume
m is mass
R is gas constant
T is temperature
We will take the volume to be constant.
So, in the initial state, we have;
P1•V = m1•R•T1 - - - eq(1)
In the final state, we have;
P2•V = m2•R•T2 - - - - eq(2)
Combining eq (1) and eq(2),we have;
P1•m2•R•T2 = P2•m1•R•T1
Dividing both sides by R gives;
P1•m2•T2 = P2•m1•T1
Making T2 the subject gives;
T2 = (P2•m1•T1)/(P1•m2)
Now, we are given;
m1 = 2kg
m2 = ½*2 = 1kg
P1 = 4 atm
P2 = 2.2 atm
T1 = 40°C = 273 + 40 K = 313K
Plugging in this values into the T2 equation, we have;
T2 = (2.2 × 2 × 313)/(4 × 1)
T2 = 344 K
Converting to °C, we have;
T2 = 344 - 273 = 71 °C
1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain
Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
A note on a piano vibrates 262 times per second . What is the period of the wave ?
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?
Explanation:
It is given that,
Total mass is 70 kg
The truck exerts a constant force of 20 N.
Then the net force is given by :
F = ma
a is acceleration of rider
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]
Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :
[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]
Since, 1 m/s = 2.23 mph
4.28 m/s = 9.57 mph
So, the speed of the rider is 4.28 m/s or 9.57 mph.
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Therefore, the distance between politician and TV set is 2536kmExplanation:
Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.
The reporter hears the sound is
4.1 / 343 = 0.01195 s later
The viewer hears the sound from the TV is
2.9 / 343 = 0.00845s
the difference is 0.00845 sec
the question is how far the TV signal can travel in that time.
the distance between politician and TV set is
= 0.00845 * 3*10^8 m
= 2536 km
d = 2536km
Therefore, the distance between politician and TV set is 2536kmUse the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, counting the ground level as the first,
A. What is the energy E of the emitted photon in electron volts?、
B. What is the wavelength in nanometers of the emitted photon?
C. What is the radius of the hydrogen atom in nanometers in its initial 5th energy level?
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex] (1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]
B. The energy of the emitted photon is given by the following formula:
[tex]E=h\frac{c}{\lambda}[/tex] (2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]
Next, you use the equation (2) and solve for λ:
[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]
C. The radius of the orbit is given by:
[tex]r_n=n^2a_o[/tex] (3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
[tex]r_5=5^2(2.380)=59.5[/tex]
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
A) The energy E of the emitted photon in electron volts is; E = 2.856 eV
B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm
C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm
A) Formula for the energy E of the emitted photons is;
E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])
We are given;
n₂ = 5
n₁ = 2
Thus;
E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])
E = 2.856 eV
B) The formula for the wavelength is;
λ = hc/E
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
E is energy of photon
λ is wavelength of the photon
Earlier we saw that E = 2.856 eV. Converting to Joules gives;
E = 4.5758 × 10⁻¹⁹ J
Thus;
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)
λ = 4.344 × 10⁻⁷ m
Converting to nm gives;
λ = 434.4nm
C) Formula for the radius of the hydrogen atom is;
rₙ = n²a₀
where;
a₀ is bohr's radius = 5.292 × 10⁻¹¹ m
n = 5
Thus;
rₙ = 5² × 5.292 × 10⁻¹¹
rₙ = 1.323 × 10⁻⁹
rₙ = 1.323 nm
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Countries create quotas and tariffs to increase the volume of trade with their neighbors.
Oooooo, that statement is not true. Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.
Answer:
False
Explanation:
I took the text :)
How can I show that the sphere of radius R performs a simple harmonic movement. how can i set its reference point and make the free body diagram.
I have the torque sum equation which is equal to the moment of inertia by angular acceleration
Explanation:
Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod). There are three forces on the pendulum:
Weight force mg at the center of the sphere,
Reaction force in the x direction at the pivot,
Reaction force in the y direction at the pivot.
Sum the torques about the pivot O.
∑τ = I d²θ/dt²
mg (L sin θ) = I d²θ/dt²
For small θ, sin θ ≈ θ.
mg L θ = I d²θ/dt²
Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.
If you wish, you can use parallel axis theorem to find the moment of inertia about O:
I = Icm + md²
I = ⅖ mr² + mL²
mg L θ = (⅖ mr² + mL²) d²θ/dt²
gL θ = (⅖ r² + L²) d²θ/dt²
Crystalline germanium (Z=32, rho=5.323 g/cm3) has a band gap of 0.66 eV. Assume the Fermi energy is half way between the valence and conduction bands. Estimate the ratio of electrons in the conduction band to those in the valence band at T = 300 K. (See eq. 10-11) Assume the width of the valence band is ΔΕV ~ 10 eV.
Answer:
= 8.2*10⁻¹²
Explanation:
Probability of finding an electron to occupy a state of energy, can be expressed by using Boltzmann distribution function
[tex]f(E) = exp(-\frac{E-E_f}{K_BT} )[/tex]
From the given data, fermi energy lies half way between valence and conduction bands, that is half of band gap energy
[tex]E_f = \frac{E_g}{2}[/tex]
Therefore,
[tex]f(E) = exp(-\frac{E-\frac{E_g}{2} }{K_BT} )[/tex]
Using boltzman distribution function to calculate the ratio of number of electrons in the conduction bands of those electrons in the valence bond is
[tex]\frac{n_{con}}{n_{val}} =\frac{exp(-\frac{[E_c-E_g/2]}{K_BT} )}{exp(-\frac{[E_v-E_fg/2}{K_BT} )}[/tex]
[tex]= exp(\frac{-(E_c-E_v}{K_BT} )\\\\=exp(\frac{-(0.66eV)}{(8.617\times10^-^5eV/K)(300K)} )\\\\=8.166\times10^-^1^2\approx8.2\times10^{-12}[/tex]
calculate the volume of marble if its diameter is 10mm
Answer:
The volume of the marble is [tex]523.33\ mm^2[/tex].
Explanation:
Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.
The volume of spherical shaped object is given by :
[tex]V=\dfrac{4}{3}\pi r^3[/tex]
Plugging all the values, we get :
[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]
So, the volume of the marble is [tex]523.33\ mm^2[/tex].
A single loop of wire with an area of 0.0820 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.220 T/s .
Required:
a. What emf is induced in this loop?
b. If the loop has a resistance of 0.600Ω, find the current induced in the loop.
Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
Learn more:
https://brainly.com/question/15477464
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
What is the relationship between electric force and distance between charged objects and the amount of charge?
Explanation:
The relationship between electric force and distance between charged objects is given by the formula as follows :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is electrostatic constant and d is distance between charges
The electric force between charges is inversely proportional to the square of distance between them.
You are helping your friend prepare for the next skateboard exhibition by determining if the planned program will work. Your friend will take a running start and then jump onto a heavy-duty 13-lb stationary skateboard. The skateboard will glide in a straight line along a short, level section of track, then up a curved concrete wall. The goal is to reach a height of at least 10 feet above the starting point before coming back down the slope. Your friend's maximum running speed to safely jump on the skateboard is 24 feet/second. Your friend weighs 155-lbs. What is the height hf that your friend will reach according to his plan?
Answer:
8.3 feet
Explanation:
The kinetic energy of the system on the ground is ...
KE = Σ(1/2)(mv^2) = (1/2)(155)(24^2) +(1/2)(13)(0^2) = 44640 lb·ft²/s²
The potential energy at the highest point is the same:
PE = mgh
44640 = (155 +13)(32)h
h = 44640/5376 = 8.30 . . . . feet
_____
We haven't worried too much about the conversion between pounds mass and pounds force. Whatever factor may be involved will divide out when computing the maximum height. We have used g=32 ft/s².
__
To achieve a 10 ft height, the running speed would need to be 26.34 ft/s, about 10% higher.
EASY HELP
As a space shuttle climbs, _____.
its mass increases
its mass decreases
its weight increases
its weight decreases
Answer: it's weight decreases
Explanation: