Answer:
Option B
Explanation:
The net emf in the circuit
E = 12 - 10 = 2 V
Total effective resistance,
r = 1 + 1 = 2 ohm
By using Ohm's law
E = I × R
I = 2 / 2 = 1 A
To find the current in the circuit, use Ohm's Law and solve two equations.
Explanation:The current in the circuit can be found using Ohm's Law: V = IR, where V is the voltage, I is the current, and R is the resistance.
In this case, the emf of the first battery is 12 V with an internal resistance of 1 Ω. Let's assume the current in the circuit is I1. So, the terminal voltage across the first battery can be calculated using Ohm's Law: 12 = I1 * (1 + 1).
Similarly, for the second battery with emf of 10 V and internal resistance of 1 Ω, the terminal voltage can be calculated as 10 = (I1 - I) * (1 + 1), where I is the current flowing in the second battery.
Now, we need to solve these two equations to find the value of I, which is the current in the circuit.
Three clowns move a 345 kg crate 12.5 m to the right across a smooth floor. Take the positive horizontal and vertical directions to be right and up, respectively. Moe pushes to the right with a force of 535 N , Larry pushes to the left with 225 N , and Curly pushes straight down with 705 N . Calculate the work ???? done by each of the clowns. Assume friction is negligible.
Answer:
Moe: 6687.5 J, Larry: -2812.5 J, Curly: 0 J
Explanation:
The work done by each clown is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force applied
d is the displacement of the box
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
Let's apply the formula to each clown:
- Moe: F = 535 N, d = 12.5 m, [tex]\theta=0^{\circ}[/tex] (because Moe pushes to the right, and the box also moves to the right)
[tex]W=(535)(12.5)cos 0^{\circ}=6687.5 J[/tex]
- Larry: F = 225 N, d = 12.5 m, [tex]\theta=180^{\circ}[/tex] (because Larry pushes to the left, while the box moves to the right)
[tex]W=(225)(12.5)cos 180^{\circ}=-2812.5 J[/tex]
- Curly: F = 705 N, d = 12.5 m, [tex]\theta=90^{\circ}[/tex] (because Curly pushes downward, while the box moves to the right)
[tex]W=(705)(12.5)cos 90^{\circ}=0[/tex]
The work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,
The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.What is work done?Work done is the force applied on a body to move it over a distance. Work done to move a body by the application of force can be given as,
[tex]W=Fd\cos\theta[/tex]
Here (F) is the magnitude of force, [tex]\theta[/tex] is the angle of force applied and (d) is the distance traveled.
The mass of the crate is 345 kg. The crate moved by the clowns is 12.5 m to the right across a smooth floor.
Moe pushes to the right with a force of 535 N-As, the Moe applies the force to the crate on the right side and the crate move in the direction of force. Here, the angle made is equal to the 0 degrees.
Therefore, the work done by the Moe is,
[tex]W=535\times12.5\times\cos (0)\\W=6687.5\rm J[/tex]
thus, the work done by the Moe is 6687.5 J.
Larry pushes to the left with 225 N. -As, the Larry applies the force to the crate on the left side and the crate move in the opposite direction of force. Here, the angle made is equal to the 180 degrees.
Therefore, the work done by the Larry is,
[tex]W=225\times12.5\times\cos (180)\\W=-2812.5\rm J[/tex]
thus, the work done by the Larry is -2812.5 J.
Curly pushes straight down with 705 N.-As, the Curly applies the force to the crate on the straight down and the crate move in the perpendicular direction of force. Here, the angle made is equal to the 90 degrees.
Therefore, the work done by the Curly is,
[tex]W=705\times12.5\times\cos (90)\\W=0\rm J[/tex]
thus, the work done by the Curly is 0 J.
Thus, the work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,
The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.Learn more about the work done here;
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A camper dives from the edge of a swimming pool at water level with a speed of 8.0 m/s at an angle of 30.0° above the horizontal. (a) How long is the diver in the air? (b) How high is the diver in the air? (c)How far out in the pool does the diver land?
Answer:
(a) 0.816 s
(b) 0.816 m
(c) 5.656 m
Explanation:
u = 8 m/s, theta = 30 degree,
(a) Use the formula for time of flight
T = 2 u Sin theta / g
T = ( 2 x 8 x Sin 30 ) / 9.8 = 0.816 s
(b) Use the formula for maximum height
H = u^2 Sin^2theta / 2 g
H = ( 8 x 8 x Sin^2 30) / ( 2 x 9.8)
H = 0.816 m
(c) Use the formula for horizontal range
R = u^2 Sin 2 theta / g
R = ( 8 x 8 x Sin 60) / 9.8
R = 5.656 m
Two man pull on a cart with a mass of 90 kg. if 1 man pulls 140 N to the right and the other man pulls 195 N to the left then find the magnitude and direction of the acceleration for the cart
Answer:
magnitude : 0.6 m/s²
Direction : Left
Explanation:
m = mass of the cart = 90 kg
Taking force in right direction as positive and force in left direction as negative
F₁ = Force applied by man 1 = 140 N
F₂ = Force applied by other man = - 195 N
a = acceleration of the cart
Force equation for the motion of the cart is given as
F₁ + F₂ = ma
140 + (- 195) = 90 a
a = - 0.6 m/s²
magnitude of acceleration is 0.6 m/s²
The negative sign indicates the direction of acceleration towards left
Final answer:
The magnitude of the cart's acceleration is approximately 0.611 m/s², and the direction is to the left.
Explanation:
When calculating the acceleration of the cart, we need to consider the net force acting on it and its mass. The net force is found by subtracting the smaller force from the larger force, taking into account their directions. With forces of 140 N to the right and 195 N to the left, the net force is the difference, which is 195 N - 140 N = 55 N, directed to the left since the larger force is in that direction. Using Newton's second law, acceleration (a) is the net force (Fnet) divided by the mass (m).
To find the magnitude of the acceleration, calculate a = Fnet / m. So, a = 55 N / 90 kg which equals approximately 0.611 m/s2. Since the larger force was to the left, the direction of the acceleration is also to the left.
A mass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. If a second mass of 3.09 kg is now added to the spring, how far with it now stretch with both masses hanging?? State your answer to the correct number of significant digits and include the proper units.
Answer:
0.694 m
Explanation:
Case 1 : When only mass of 2.82 kg is hanged from spring
m = mass hanged from the spring = 2.82 kg
x = stretch caused in the spring = 0.331 m
k = spring constant
Using equilibrium of force in vertical direction
Spring force = weight of the mass
k x = m g
k (0.331) = (2.82) (9.8)
k = 83.5 N/m
Case 2 : When both masses are hanged from spring
m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg
x = stretch caused in the spring = ?
k = spring constant = 83.5 N/m
Using equilibrium of force in vertical direction
Spring force = weight of the mass
k x = m g
(83.5) x = (5.91) (9.8)
x = 0.694 m
A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (μ 0 = 4π × 10-7 T · m/A
Answer:
104
Explanation:
U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J
i = current flowing through the solenoid = 0.4 A
L = inductance of the solenoid
Energy stored in the solenoid is given as
U = (0.5) L i²
6.00 x 10⁻⁶ = (0.5) L (0.4)²
L = 75 x 10⁻⁶
Inductance is given as
l = length of the solenoid = 0.7 m
N = number of turns
r = radius = 5.00 cm = 0.05 m
Area of cross-section is given as
A = πr²
A = (3.14) (0.05)²
A = 0.00785 m²
Inductance is given as
[tex]L=\frac{\mu _{o}N^{2}A}{l}[/tex]
[tex]75\times 10^{-6}=\frac{(12.56\times 10^{-7})N^{2}(0.00785)}{0.7}[/tex]
N = 73
Winding density is given as
density = n = [tex]\frac{N}{l}[/tex]
n = [tex]\frac{73}{0.7}[/tex]
n = 104
A coil has a resistance of 5Ω and an inductance of 100 mH . At a particular instant in time after a battery is connected across the coil, the current is 2 A, and is increasing at a rate of 20A/s a) What is the voltage of the battery. b) What is the time-constant of the circuit. c) What is the final value of the current. d) Plot I() Vs time.
I have seen this question before and the correct answer would be B
Hope this helped!!
Answer:
The correct answer would be B
Explanation:
A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at one end. The resistance of the bar and the rails is constant and equal to 0.10 Ω. A uniform magnetic field is perpendicular to the plane of the rails. A 0.080-N force parallel to the rails is required to keep the bar moving at a constant speed of 0.50 m/s. What is the magnitude of the magnetic field?
Answer:
0.25 T
Explanation:
F = Force required to keep the bar moving = 0.080 N
B = magnitude of magnetic field = ?
L = length of the bar = 50 cm = 0.50 m
v = speed of the bar = 0.50 m/s
R = resistance of the bar =0.10 Ω
Force is given as
[tex]F = \frac{B^{2}L^{2}v}{R}[/tex]
[tex]0.08 = \frac{B^{2}(0.50)^{2}(0.50)}{0.10}[/tex]
B = 0.25 T
To find the magnitude of the magnetic field, use the equation F = qvBsinθ and Ohm's Law to solve for q. Then substitute q back into the equation to find the magnetic field.
Explanation:To find the magnitude of the magnetic field, we can use the equation F = qvBsinθ. In this case, the force (F) is given as 0.080 N, the charge (q) is not given but is not needed to find the magnetic field, the velocity (v) is given as 0.50 m/s, and sinθ is 1 because the angle between the velocity and the magnetic field is 90°. Rearranging the equation to solve for B, we get B = F / (qv sinθ). Substituting the given values, we have B = 0.080 N / (q * 0.50 m/s * 1). The only unknown variable left is the charge (q).
To calculate the charge, we can use the given resistance and Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Rearranging the equation to solve for I, we get I = V / R. The voltage across the resistor can be calculated using the formula V = Fd, where d is the distance between the rails, which is given as 50 cm (or 0.5 m). Substituting the given values, we have V = 0.080 N * 0.5 m. Now, we can substitute the calculated current (I) into the equation B = 0.080 N / (q * 0.50 m/s * 1) and solve for q. Finally, we can substitute the calculated charge (q) back into the equation B = 0.080 N / (q * 0.50 m/s * 1) to find the magnitude of the magnetic field.
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Problem 1: A positively charged particle Q1 = +35 nC is held fixed at the origin. A second charge Q2 of mass m = 8.5 micrograms is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. A) What is the sign of charge on Q2? B) Calculate the magnitude of Q2 in units of nanocoulombs.
Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)
Explanation:
The sign of charge Q2 is positive, and its magnitude can be calculated by setting the gravitational force equal to the electric force due to Q1 and solving for Q2.
Explanation:To determine the sign and magnitude of the second charge Q2, we use the concept that the net force acting on it must be zero. The forces that act on Q2 include the gravitational force and the electric force due to Q1. Since Q1 is positive and the net force is zero, Q2 must also be positive for the electric force (Coulomb force) to balance the gravitational pull downwards.
The gravitational force can be calculated using Newtons's law of gravitation F = mg, where g is the acceleration due to gravity (9.8 m/s2). The electric force is given by Coulomb's law F = k|Q1Q2|/d2, where k is Coulomb's constant (8.99 x 109 Nm2/C2).
Setting the magnitude of the gravitational force equal to the magnitude of the electric force gives the equation mg = k|Q1Q2|/d2. Solving for Q2, and converting the mass of Q2 from micrograms to kilograms, we calculate the magnitude of charge Q2.
Determine explicitly which is faster, 75 miles per hour or 75 m/s? Express all your results in MKS units and explain your conclusions from the numbers, and explain using words. Show all work.
Answer:
75 m/s is faster
Explanation:
MKS units stands for meter kilogram seconds
75 miles per hour = 75 mph
1 mile = 1609.34 meters
1 hour = 60×60 = 3600 seconds
1 mph = 1609.34/3600 = 0.44704 m/s
75 mph = 75×0.44707 = 33.52792 m/s
Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x 5 5.00 cm and released from rest at t 5 0. (a) What is the force constant of the spring? (b) What are the angular frequency v, the frequency, and the period of the motion? (c) What is the total energy of the system?
Answer:
a) 250 N/m
b) 22.4 rad/s , 3.6 Hz , 0.28 sec
c) 0.3125 J
Explanation:
a)
F = force applied on the spring = 7.50 N
x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m
k = spring constant of the spring
Since the force applied causes the spring to stretch
F = k x
7.50 = k (0.03)
k = 250 N/m
b)
m = mass of the particle attached to the spring = 0.500 kg
Angular frequency of motion is given as
[tex]w = \sqrt{\frac{k}{m}}[/tex]
[tex]w = \sqrt{\frac{250}{0.5}}[/tex]
[tex]w [/tex] = 22.4 rad/s
[tex]f[/tex] = frequency
Angular frequency is also given as
[tex]w [/tex] = 2 π [tex]f[/tex]
22.4 = 2 (3.14) f
[tex]f[/tex] = 3.6 Hz
[tex]T[/tex] = Time period
Time period is given as
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{3.6}[/tex]
[tex]T[/tex] = 0.28 sec
c)
A = amplitude of motion = 5 cm = 0.05 m
Total energy of the spring-block system is given as
U = (0.5) k A²
U = (0.5) (250) (0.05)²
U = 0.3125 J
(a) The force constant of the spring is 250 N/m.
(b) The angular frequency of the mass oscillation is 22.36 rad/s, frequency is 3.56 Hz and the period is 0.28 s.
(c) the total energy of the system is 0.31 J.
Force constant of the springThe force constant of the spring can be determined by applying Hooke's law as follows;
F = kx
k = F/x
k = (7.5)/0.03)
k = 250 N/m
Angular frequencyThe angular frequency of the mass oscillation is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{250}{0.5} }\\\\\omega = 22.36 \ rad/s[/tex]
Angular frequencyThe angular frequency is calculated as follows;
ω = 2πf
f = ω/2π
f = (22.36)/2π
f = 3.56 Hz
Period of the oscillationThe period of the oscillation is calculated as follows;
T = 1/f
T = 1/3.56
T = 0.28 s
Total energy of the systemThe total energy of the system is calculated as follows;
U = ¹/₂kA²
U = ¹/₂ x 250 x (0.05)²
U = 0.31 J
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A golf ball was thrown with a speed of 25 m/sec at an angle of 65° above the horizontal. What is the horizontal distance traveled by the ball before it hits the ground?
Answer:
The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.
Explanation:
It is given that,
Speed of golf ball, v = 25 m/s
Angle above horizontal or angle of projection, θ = 65°
We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.
[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]
[tex]R=\dfrac{(25\ m/s)^2\ sin2(65)}{9.8\ m/s^2}[/tex]
R = 48.85 m
So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.
A DVD drive is spinning at 100.0 rpm. A dime (2.00 gm) is placed 3.00 cm from the center of the DVD. What must the coefficient of friction be to keep the dime on the disk?
Answer:
0.3375
Explanation:
w = angular speed of the DVD drive = 100.0 rpm = [tex]100.0 \frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex] = [tex]10.5\frac{rad}{sec}[/tex]
m = mass of the dime = 2 g = 0.002 kg
r = radius = 3 cm = 0.03 m
μ = Coefficient of friction
The frictional force provides the necessary centripetal force to move in circle. hence
frictional force = centripetal force
μ mg = m r w²
μ g = r w²
μ (9.8) = (0.03) (10.5)²
μ = 0.3375
A 79-kg worker at the Burj Khalifa, the tallest building in the world with its roof 828 m above the ground below, rides a series of elevators from the ground level to the roof. What is the worker's change in gravitational potential energy?
Answer:
Worker's change in gravitational potential energy = 641.69 kJ
Explanation:
Potential energy = Mass x Acceleration due to gravity x Height
PE = mgh
Mass, m = 79 kg
Acceleration due to gravity, g = 9.81 m/s²
Height, h = 828 m
Potential energy, PE = 79 x 9.81 x 828 = 641691.72 J = 641.69 kJ
Worker's change in gravitational potential energy = 641.69 kJ
The Sun radiates electromagnetic energy at the rate of 3.92 × 1026 W. The mass of the Sun is 1.99 × 1030 kg. What fraction of the Sun’s mass is lost during a human lifetime of 75 years? A. 2.16 × 10–13 B. 1.42 × 10–14 C. 6.90 × 10–14 D. 5.18 × 10–12 E. 8.63 × 10
Answer:
D. 5.18 x 10⁻¹²
Explanation:
[tex]\frac{dE}{dt}[/tex] = rate at which sun radiates energy = 3.92 x 10²⁶ W
M = mass of sun = 1.99 x 10³⁰ kg
[tex]\frac{dm}{dt}[/tex] = rate at which sun's mass is lost
c = speed of light
Energy is given as
E = m c²
Taking derivative both side relative to "t"
[tex]\frac{dE}{dt}=c^{2}\frac{dm}{dt}[/tex]
[tex]3.92\times 10^{26}=(3\times 10^{8})^{2}\frac{dm}{dt}[/tex]
[tex]\frac{dm}{dt}[/tex] = 4.4 x 10⁹ kg/s
t = time interval = 75 yrs = 75 x 365 days = 75 x 365 x 24 hours = 75 x 365 x 24 x 3600 sec = 2.4 x 10⁹ sec
[tex]m[/tex] = mass lost
mass lost is given as
[tex]m = t\frac{dm}{dt}[/tex]
[tex]m = (2.4\times 10^{9})(4.4\times 10^{9})[/tex]
m = 10.56 x 10¹⁸ kg
fraction is given as
fraction = [tex]\frac{m}{M}[/tex]
fraction = [tex]\frac{10.56\times 10^{18}}{1.99\times 10^{30}}[/tex]
fraction = 5.18 x 10⁻¹²
Water is moving at a velocity of 2.17 m/s through a hose with an internal diameter of 1.60 cm. The fluid velocity in this hose’s nozzle is 14.5 m/s. What is the nozzle’s inside diameter, in cm?
Final answer:
To find the nozzle's inside diameter, use the principle of continuity. Set up the equation A1V1 = A2V2 and solve for r2. Finally, calculate the nozzle's inside diameter.
Explanation:
To find the nozzle's inside diameter, we can use the principle of continuity, which states that the flow rate of a fluid is constant throughout a pipe or nozzle.
The formula for continuity is A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity.
In this case, we can set up the equation as (pi * r12) * 2.17 = (pi * r22) * 14.5, where r1 and r2 are the radii of the hose and nozzle respectively.
By rearranging the equation and solving for r2, we find that r2 = sqrt((r12 * 2.17) / 14.5).
Since the radius is half of the diameter, the nozzle's inside diameter can be calculated as 2 * r2 in cm.
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.00 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.5 cm and it takes 0.45 s to move it into the field. What average current is induced in the ring in A if its resistance is 0.0100 Ω ?
The average induced current in the loop is 0.218 A.
Induced emf in the loopThe emf induced in the loop is determined by applying Faraday's law as shown below;
emf = dФ/dt
emf = BA/t
where;
A is the area
A = πr² = πd²/4
A = π x (0.025)²/4
A = 4.908 x 10⁻³ m²
emf = (2 x 4.908 x 10⁻³)/(0.45)
emf = 2.18 x 10⁻³ V
Average induced currentThe average induced current in the loop is calculated as follows;
I = emf/R
I = 2.18 x 10⁻³/0.01
I = 0.218 A
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A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m
Answer:
The highest of its trajectory = 0.45 m
Option C is the correct answer.
Explanation:
Considering vertical motion of cat:-
Initial velocity, u = 3.44 sin60 = 2.98 m/s
Acceleration , a = -9.81 m/s²
Final velocity, v = 0 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0² = 2.98² + 2 x -9.81 x s
s = 0.45 m
The highest of its trajectory = 0.45 m
Option C is the correct answer.
A moon is in orbit around a planet. The moon's orbit has a semimajor axis of 4.3 times 10 Superscript 8 Baseline m and has an orbital period of 1.516 days. Use these data to estimate the mass of the planet.
Answer:
The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].
Explanation:
Given that,
Semi major axis [tex]a= 4.3\times10^{8}[/tex]
Orbital period T=1.516 days
Using Kepler's third law
[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]
[tex]M=\dfrac{4\pi^2}{GT^2}a^3[/tex]
Where, T = days
G = gravitational constant
a = semi major axis
Put the value into the formula
[tex]M=\dfrac{4\times(3.14)^2}{6.67\times10^{-11}(1.516\times24\times60\times60)^2}(4.3\times10^{8})^3[/tex]
[tex]M=2.7\times10^{27}\ kg[/tex]
Hence, The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].
A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle between the electric field vector and the vector normal to the rectangular plane is 65.0°. What is the electric flux through the rectangle?
Answer:
6.6 x 10^5 Nm^2/C
Explanation:
E = 125000 N/C
Area, A = length x width = 2.5 x 5 = 12.5 m^2
θ = 65 degree
electric flux, φ = E A Cosθ
φ = 125000 x 12.5 x Cos 65
φ = 6.6 x 10^5 Nm^2/C
Here, we are required to evaluate the electric flux through the rectangle.
The electric flux through the rectangle is;
φ = 6.6 × 10^5 Nm²/C
The electric flux through a body is given as;
electric flux, φ = E A Cosθ
where;.
φ = electric fluxE = Electric field magnitudeA = Areaθ = angle between the electric field vector and the vector normalThe Area of the rectangle, A = 2.5 × 5
Therefore, A = 12.5m².
Therefore, the electric flux, φ = E A Cosθ
φ = 125000 × 12.5 × Cos65
φ = 660,341 Nm²/C.
φ = 6.6 × 10^5 Nm²/C
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A hydrogen atom in the ground state absorbs a photon that increases its principle quantum number by 3. How many different wavelengths could the atom emit on the decay path back to the ground state? (Ignore spin-orbit coupling for this question).
Answer:
[tex]N = 3[/tex]
so these are
either 3 to 2 or 2 to 1 or 3 to 1
Explanation:
As we know that total number photons emitted from nth state to ground state is given by
[tex]N = ^nC_2[/tex]
[tex]N = \frac{n(n-1)}{2}[/tex]
here it shows that if electron has transition from nth excited state to ground state then total number of photons is given by above equation
here we know that higher state is n = 3
so total number of photons is given as
[tex]N = \frac{3(3-1)}{2}[/tex]
[tex]N = 3[/tex]
so these are
either 3 to 2 or 2 to 1 or 3 to 1
A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 3.3 kg, and its initial speed is 3.2 m/s. How much does the block's temperature increase, if it absorbs 74% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J/(kg · °C).
°C
Answer:
Rise in temperature is given as
[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]
Explanation:
Initial kinetic energy of the block is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
here we will have
m = 3.3 kg
v = 3.2 m/s
now we will have
[tex]KE = \frac{1}{2}mv^2[/tex]
now we will have
[tex]KE = \frac{1}{2}(3.3)(3.2)^2[/tex]
[tex]KE = 17 J[/tex]
now we know that 74% of initial kinetic energy is absorbed as internal energy of the block
so the rise in temperature of the block is given as
[tex]KE = ms\Delta T[/tex]
[tex]0.74 \times 17 J = (3.3)(452)\Delta T[/tex]
[tex]12.5 = 1491.6 \Delta T[/tex]
[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]
Answer:
The temperature increases 0.0084ºC
Explanation:
Please look at the solution in the attached Word file.
A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. Is the change of the boulder’s momentum in one second greater than, less than, or equal to the change of the pebble’s momentum in the same time period?
Answer:
equal
Explanation:
According to the Newton's second law of motion, the force acting on the body is directly proportional to the rate of change of momentum.
If the force is same, then the change in momentum is also same for a given time.
The change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration.
The given data in the problem is;
F is the net force=200 N
m is the mass of boulder=100 kg
M is the mass of pebble=100 g
The force acting on the body is directly proportional to the rate of change in momentum, according to Newton's second law of motion.
For a given time, if the force is the same, the change in momentum is likewise the same.
Hence the change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.
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Polarizing Filters In the fifth lab (the ters and measured the resulting intensity. Initially vertically polarized light, with an intensity Io, is sent through two (ideal) polarizing sheets, whose transmission axes are 45° and 90° to the vertical. Will the intensity of the light be greater or less than if with the polarizers), you shined light through polarizhng there were only polarizer at 90° to the yertical? Explain. Note that Malus Law one says I2= I cos2 e
Answer:
So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical
Explanation:
When two polarizers are arranged at 45 degree and 90 degree in series
so here the intensity of light coming out of the polarizers is given as
[tex]I_1 = I_ocos^45[/tex]
[tex]I_1 = \frac{I_0}{2}[/tex]
now for second polarizer we have
[tex]I_2 = I_1 cos^2(90 - 45)[/tex]
[tex]I_2 = (\frac{I_0}{2})\frac{1}{2}[/tex]
[tex]I_2 = \frac{I_0}{4}[/tex]
now in other case when two polarizers are inclined 90 degree to the vertical
[tex]I = I_ocos^290 = 0[/tex]
so final intensity in second case will be ZERO
So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical
A large building shaped like a box is 40 m high with a face that is 6565 m wide. A strong wind blows directly at the face of the building, exerting a pressure of 170 N/m2 at the ground and increasing with height according to P(y)=170+2y, where y is the height above the ground. Calculate the total force on the building, which is a measure of the resistance that must be included in the design of the building.
Answer:
[tex]F=55146000 N[/tex]
Explanation:
Given:
Pressure on the building = 170 N/m²
Pressure function P(y) = 170 + 2y
Consider a small height of the building as dy
the small force (dF) exerted on the height dy will be given as
dF = P(y)×(Area of the building face)
or
dF = (170 + 2y)(6565)dy
or
dF = 170×6565×dy + 2y×6565×dy
dF = 1116050dy + 13130ydy
integerating the function for the whole height of 40 m
[tex]\int\limits^{40}_0 {F} = \int _0^{40}\:\left(\:1116050dy\:+\:13130ydy\right)[/tex]
[tex]F = \int _0^{40}1116050dy+\int _0^{40}13130ydy[/tex]
[tex]F=\left[1116050y\right]^{40}_0+13130\left[\frac{y^2}{2}\right]^{40}_0[/tex]
[tex]F=44642000+10504000[/tex]
[tex]F=55146000 N[/tex] ; Total force exerted on the face of the building
To calculate the total force on a building due to wind pressure, integrate the pressure function over the building's height and multiply by the building's width. This calculation is vital for determining the building's required resistance to wind forces for safe and stable design.
Explanation:The question asks to calculate the total force exerted on a large building by a wind blowing against one of its faces. The wind pressure increases with height as described by the function P(y)=170+2y, where y is the height in meters above the ground. To find the total force, we need to integrate this pressure function over the height of the building and multiply by the width of the building's face.
First, we can write the integral of the pressure over the height of the building:
Integrate [P(y) dy] from y=0 to y=40
= Integrate [(170+2y) dy] from y=0 to y=40
Performing this integration will give us the total pressure exerted over the entire height of the building. After integrating, we multiply the result by the width of the building (6565 m) to obtain the total force.
Using the integration results, the total force, F, can be expressed as:
F = (Integral result) × Width of the building
This force calculation is critical for architects and engineers as it provides an estimate of the resistance that the building must be designed to withstand due to wind forces, ensuring structural stability and safety.
A water pump draws about 9 A when connected to 240 V. What is the cost (with electrical energy at 13 cents per kWh) of running the pump for 16 h? (Give your answer in dollars)
Answer:
4.49 dollars
Explanation:
i = 9 A, V = 240 V, t = 16 h
Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh
The cost of 1 kWh is 13 cents.
Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars
The cost of running a water pump drawing 9 A at 240 V for 16 hours, with an energy cost of 13 cents per kWh, is $4.49.
Explanation:To calculate the cost of running a water pump for 16 hours at an electrical energy cost of 13 cents per kWh, you need to follow these steps:
First, calculate the power consumed by the pump. Power (P) is given by the formula P = VI, where V is the voltage and I is the current. For a pump drawing 9 A at 240 V, the power consumed is P = 240V * 9A = 2160W or 2.16kW.Next, calculate the total energy consumed in kilowatt-hours. Energy (E) is calculated as E = Pt, where P is the power in kW and t is the time in hours. For 16 hours, the energy consumed is E = 2.16kW * 16h = 34.56kWh.Finally, calculate the cost by multiplying the total energy consumed by the cost per kilowatt-hour. The cost (C) is C = E * cost/kWh. So, the cost to run the pump for 16 hours is C = 34.56kWh * $0.13/kWh = $4.4928.In dollars, the cost of running the pump for 16 hours is $4.49 (rounded to the nearest cent).
An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of the rotor? O 305,200 m/s O 152.3 m/s O 15.20 m/s O 3042 m/s
Answer:
Option D is the correct answer.
Explanation:
Since value of angular acceleration is constant, the body has only centripetal acceleration.
Centripetal acceleration
[tex]a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2[/tex]
We have radius = 7.112 cm = 0.07112 m
Frequency, f = 1975 rpm = 32.92 rps
Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s
Substituting in centripetal acceleration equation,
[tex]a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2[/tex]
Option D is the correct answer.
If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1 R3 . If the resistances are measured in ohms as R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R. (Round your answer to three decimal places.) Ω
To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.
Explanation:To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.
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The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.
Explanation:The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:
1/R = 1/R1 + 1/R2 + 1/R3
To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.
After calculating 1/R using the given resistances:
1/R = 1/100 + 1/25 + 1/10
1/R = 0.01 + 0.04 + 0.1 = 0.15
Therefore, R = 1 / 0.15 = 6.667 Ω
We calculate the maximum possible errors in resistances as:
Error in R1 (eR1) = 0.005 * 100 Ω = 0.5 ΩError in R2 (eR2) = 0.005 * 25 Ω = 0.125 ΩError in R3 (eR3) = 0.005 * 10 Ω = 0.05 ΩUsing the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:
eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |
Plugging in the values:
eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |
eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)
The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.
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The acceleration of high-speed aircraft is sometimes expressed in g's (in multiples of the standard acceleration of gravity). Determine the upward force, in N, that a 78.03-kg person would experience in an aircraft whose acceleration is 5.05 g's. Assume the standard acceleration of gravity is 8.23 m/s^2.
Answer:
3888.23 N
Explanation:
m = mass of the person = 78.03 kg
g = standard acceleration due to gravity = 8.23 m/s²
a = acceleration of the aircraft = 5.05 g = 5.05 x 8.23 = 41.6 m/s²
F = upward force the person experience
Force equation for the motion of person is given as
F - mg = ma
Inserting the values
F - (78.03)(8.23) = (78.03)(41.6)
F = 3888.23 N
A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the person and the kayak is 75 kg, what is the magnitude of the net force acting on the kayak?
Answer:
Magnitude of the net force acting on the kayak = 39.61 N
Explanation:
Considering motion of kayak:-
Initial velocity, u = 0 m/s
Distance , s = 0.40 m
Final velocity, v = 0.65 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0.65² = 0² + 2 x a x 0.4
a = 0.53 m/s²
We have force, F = ma
Mass, m = 75 kg
F = ma = 75 x 0.53 = 39.61 N
Magnitude of the net force acting on the kayak = 39.61 N
Calculate the centripetal force on a 6.0 kg sensor that is attached at the end of a 100.0-m a long wind turbine blade that is rotating at 1.25 rev/s.
Answer:
Centripetal force, F = 18486.75 N
Explanation:
It is given that,
Mass of the sensor, m = 6 kg
It is attached at the end of 100 m long wind turbine, d = 100 m
So, the radius of the wind turbine, r = 50 m
Angular velocity, [tex]\omega=1.25\ rev/s=7.85\ rad/s[/tex]
The centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
Since, [tex]v=r\omega[/tex]
[tex]F=mr\omega^2[/tex]
[tex]F=6\ kg\times 50\ m\times (7.85\ rad/s)^2[/tex]
F = 18486.75 N
So, the centripetal force is 18486.75 N. Hence, this is the required solution.