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Yuri [45]
3 years ago
7

The distance between a charge and the source of an electric field changes from 3 mm to 6 mm.

Physics
2 answers:
V125BC [204]3 years ago
8 0

Answer:

c

Explanation:

Solnce55 [7]3 years ago
6 0

Answer:

multiplied by 2

Explanation:

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A(n) 1946 kg car travels at a speed of 10 m/s . What is its kinetic energy ? Answer in units of J.
erica [24]

Answer:

KE=97300J

Explanation:

KE=1/2mv^2

KE=1/2(1946)(10)^2

KE=97300J

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3 years ago
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Answer:

Beta 17,000K, bc warmer is more blue

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A suitcase is pulled 28 feet along a flat sidewalk with a constant upward force of 80 lb at an angle of 23degrees with the horiz
Elenna [48]

Answer:

2795.3 J

Explanation:

distance, d = 28 feet = 8.53 m

Force, F = 80 lb = 356 N

Angle, θ = 23°

Work = F x d x cos θ

W = 356 x 8.53 x cos 23

W = 2795.3 J

Thus, the work done is 2795.3 J.

5 0
4 years ago
Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr
aivan3 [116]

Answer:

The anser is b

Explanation:

4 0
3 years ago
A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to
Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

\tau = I\alpha

Where \tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

But, the angular acceleration is given by

\alpha = \frac{\omega}{t}

Where \omega is the angular speed

and t is time

Then, we can write that

\tau = \frac{I\omega}{t}

Hence,

\omega = \frac{\tau t}{I}

Now, to determine the angular speed, we first determine the Torque \tau and the moment of inertia I.

Here, The torque is given by,

\tau = rF

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ \tau = 3.00 \times 195

\tau = 585 Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

I = \frac{1}{2}MR^{2}

Where M is the mass and

R is the radius

∴I = \frac{1}{2} \times 325 \times (3.00)^{2}

I = 1462.5 kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

\omega = \frac{\tau t}{I}

\omega = \frac{585 \times 2.05}{1462.5}

\omega = 0.82 rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0
3 years ago
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