What is any motion in which an object is moving along a curved path

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a

anyanavicka [17]

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

$F_2-F_1-F_f=Ma$ (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

$a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}$

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero

3 0

3 years ago

Help please thank you

trasher [3.6K]

I believe the answer would be C because potential energy is affected by height and mass. The truck in photo C is the highest and has a lot of mass.

4 0

3 years ago

Rutherford hypothesized that, in his experiment,more alpha particles would be deflected if ?

Alex17521 [72]

This is the CBS's this is a B a bit

5 0

3 years ago

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

$\tau_{net} = \tau + \tau'$

where

$\tau$ = Torque due to Force, F

$\tau'$ = Torque due to Force, F'

$tau = F\times r$

$tau' = F'\times r'$

Now,

$\tau_{net} = - F\times r + F'\times r'$

$\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm$

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

$|\tau_{net}| = 2.41\ Nm$

3 0

3 years ago

In what state of matter are particles vibrating and moving at high speeds?

VARVARA [1.3K]

Answer:

Gas

Explanation:

6 0

3 years ago