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I am Lyosha [343]
2 years ago
4

The power supplied to a typical black-and-white television set is 90 W when the set is connected to 120 V. (a) How much electric

al energy does this set consume in 1 hour? (b) A color television set draws about 2.5 A when connected to 120 V. How much time is required for it to consume the same energy as the black-andwhite model consumes in 1 hour?
Physics
1 answer:
NeTakaya2 years ago
5 0

Answer:

The electrical energy and time required for it to consume the same energy are 3.24\times10^{5}\ J and 18 min.

Explanation:

Given that,

Power = 90 W

Voltage V = 120 V

Current = 2.5

(a). We need to calculate the  amount of energy

Using formula of energy

Q=pt

Where, p = power

t = time

Put the value into the formula

Q=90\times1

Q=90\times3600

Q=3.24\times10^{5}\ J

(b). We need to calculate the  power

Using formula of power

P=V\times I

Where, I = current

V = voltage

Put the value into the formula

P=120\times2.5

P=300\ W

We need to calculate the time

Using formula of time

t= \dfrac{Q}{P}

Put the value into the formula

t=\dfrac{3.24\times10^{5}}{300}

t=1080\s

t=\dfrac{1080}{60}=18\ min

Hence, The electrical energy and time required for it to consume the same energy are 3.24\times10^{5}\ J and 18 min.

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This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating
Olegator [25]

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

3 0
2 years ago
Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s
Black_prince [1.1K]

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

1.91J = I/2*\omega^2

where I = 2/3*m*R^2=0.23kg.m^2

Solving for ω:

\omega = \sqrt{W*2/I} =4.07rad/s

7 0
2 years ago
A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction
aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0
2 years ago
A car wheel turns through 277° in 10.7 s. Calculate the angular speed of the wheel.
slava [35]

Answer:

The angular speed of the wheel is 0.452 rad/s

Explanation:

The angle through which the car wheel turns, Δθ = 277° = 277/360 × 2·π radian

The time it takes for the car wheel to turn, Δt = 10.7 s

The angular speed, ω is given by the following equation;

Angular \ speed = \dfrac{Change \ in \ angular \ rotation }{Change \ in \ time} = \dfrac{\Delta \theta}{\Delta t}

Substituting the known values for Δθ and Δt gives;

Angular \ speed = \dfrac{\dfrac{277 ^{\circ}}{360 ^{\circ }  }  \times 2 \times \pi \ radian}{10.7 \ seconds} \approx 0.452 \ rad/s

The angular speed of the wheel = 0.452 rad/s

3 0
2 years ago
The force on a wire is a maximum of6.71 10-2 N when placed between the pole faces of a magnet.The current flows horizontally to
Taya2010 [7]

Answer:

B.   i=2.79A

C.   F=0.066N

Explanation:

A) By the right hand rule we have that

F=iL x B

F=iLBsin(α)

If the wire jump toward the observer the top pole face is the magnetic southpole.

B) The diameter of the pole face is 15cm. We can take this value as L (the length in which the wire perceives the magnetic field). Hence, we have

F=iLBsin(\alpha)\\\alpha=90°\\F=iLB\\i=\frac{F}{LB}=\frac{6.71*10^{-2}N}{(0.15m)(0.16T)}=2.79A

C) Now the length of the wire that feels B is

L=\frac{0.15m}{cos(10\°)}=0.152m

and the force will be (by taking the degrees between the magnetic field vector and current vector as 80°)

F=iLBsin(\alpha)\\F=(2.79A)(0.152m)(0.16T)(sin(80\°))=0.066N

I hope this is useful for you

regards

8 0
2 years ago
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