Is the electric potential energy of a particle with charge q the same at all points on an equipotential surface?

A compact car has a maximum acceleration of 4.0 m/s2 when it carries only the driver and has a total mass of 1200 kg . you may w

inna [987]

<span>Let F represent the maximum thrust produced by the car's motor. Thus, F = ma = 1300 x 3.0 = 3900 N. After adding the load, F stays the same, leading to the equation F = 1700a, which results in a = F/1700 = 3900/1700 = 2.3 m/s².</span>

3 0

13 days ago

Read 2 more answers

Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart

Keith_Richards [1034]

Answer: $F=\frac{(M+m)g}{\mu _s}$

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is $\mu _s$ .

A force F is applied to mass m.

The acceleration of the system is

$a=\frac{F}{M+m}$

The frictional force will counterbalance the weight of the block.

The frictional force is $=\mu _sN$

$N=ma$

$\mu _sN=mg$

$\mu _sma=mg$

$\mu _s=\frac{g}{a}$

$F=\frac{(M+m)g}{\mu _s}$

6 0

4 days ago

Read 2 more answers

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

Yuliya22 [1153]

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0

9 days ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

serg [1198]

The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.

8 0

4 days ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

ValentinkaMS [1149]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is $\frac{4}{3} \pi (\frac{D}{2})^3$