Answer: [tex]2m/s^2[/tex]
Explanation:
[tex]Formula: F=ma[/tex]
Where;
F = force
m = mass
a = acceleration
Solve for a;
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
b) The length of the rod:
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]
Given:
d = 1.5 mλ = 2.5 nC/m
Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.
Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]
How to calculate Electric Field?E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]
Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis
[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.
Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]
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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
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 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.
"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
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.
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 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]
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 .
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
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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Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now
Answer:
1.92 Watt lost
Explanation:
Power rating of each toaster = 600 Watts
Current that flows = 5 Amperes
Wasted power = 1 Watt
Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R
where I = current
and R = Resistance
600 = [tex]5^{2}[/tex] x R
R = 600/25 = 24 Ohms.
According to joules loss due to heating of wire
Power loss P ∝ [tex]I^{2}[/tex]R
imputing values,
1 ∝ [tex]5^{2}[/tex] x 24
1 ∝ 600
to remove the proportionality sign, we introduce a constant k
1 = 600k
k = 1/600 = 0.00167
For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.
The resistance across the wire becomes
1200 = [tex]10^{2}[/tex]R
R = 1200/100 = 12 Ohms
power loss P = k x [tex]I^{2}[/tex]R
P = 0.0016 x [tex]10^{2}[/tex] x 12
P = 1.92 Watt lost
This question involves the concepts of power, current, and resistance.
The power wasted by the wires in the home for two units will be "4 watt".
POWER WASTAGEThe power wasted by the wires can be given in terms of current and resistance by the following formula:
[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]
where,
P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 ATherefore,
[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]
P₂ = 4 watt
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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
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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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.
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]
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
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]
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.
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:
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
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
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?
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
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
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
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A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?
Explanation:
We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of the pendulum
g is acceleration due to gravity on which it is placed
It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.
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]
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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