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4 years ago

What is the car is acceleration if it increases its speed from 5 meters per second to 20 m/s in three seconds

1 answer:

7 0

Initial Velocity, u=5m/s
Final Velocity, v=20m/s
Time, t=3s
Then we have acceleration, a=(v-u)÷t
a=(20-5)÷3=15÷3=5

The final answer is 5m/s²

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A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the c

kaheart [24]

Answer:

787528.7 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).

From the question,

W = Tcos∅(d)............. Equation 1

Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.

But,

d = v(t)..................... equation 2

Where v = velocity, t = time

Substitute equation 2 into equation 1

W = Tcos∅(vt)............. Equation 3

Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s

Substitute into equation 3

W = 210(cos23°)(1.94×2100)

W = 787528.7 J

8 0

3 years ago

Name two important differences between light waves and sound waves

emmasim [6.3K]

Answer:

1. Light is electromagnetic waves while sound is mechanical.

Light wave is transverse and sound is longitudinal.

Explanation:

<i = <r

when distance between source and obstacle and distance between obstacle and screen is finite then the diffraction is fresnal diffraction

when distance between source and obstacle and distance between obstacle and screen is infinite then the diffraction is Fraunhofer's diffraction

8 0

3 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.