What will happen if the two plungers are pressed together firmly? Explain your answer.

Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m

marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 $m / s^{2}$

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

$F_{n e t}=m_{t o t} \times a$

Where,

$F_{n e t}$ is net force acting on body

$m_{\mathrm{tot}}$ is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

$a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}$

Put the value for m = hanging mass = 40 kg and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$ , we get

$a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}$

The tension in the ropes, $T=(m \times g)+(m \times a)$

Here, m as hanging mass

T = tension, N or $k g m / s^{2}$

m = mass, kg

g = gravitational force, $9.8 \mathrm{m} / \mathrm{s}^{2}$

a = acceleration, $m / s^{2}$

$T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N$

3 0

3 years ago

A shopper is pushing a cart down a grocery store aisle. Starting from rest, the shopper applies a constant force to the cart for

inysia [295]

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

A shopper is pushing a cart down a grocery store aisle. The movement of the cart is:

It starts from rest.
From t = 0 s to t = 4.0 s it is accelerated with a constant force.
From t = 4 s to t = 8.0 s it receives just enough force to balance the friction on the cart.
From t = 8 s to t = 12 s it is decelerated until it comes to rest.

All in all, at the initial time (t = 0 s), the velocity is 0 m/s (rest) and at the final time (t = 12 s) the velocity is 0 m/s as well (rest). The average acceleration in that period is:

$a = \frac{v_{12}-v__o}{t_{12}-t_0} = \frac{0m/m-0m/s}{12s-0s} = 0 m/s^{2}$

A shopping cart that starts from rest, is accelerated for 4 s, moves at constant velocity for 4 s, and is decelerated for 4s until returning to rest, has an average acceleration of 0 m/s².

Learn more: brainly.com/question/16274121

3 0

2 years ago

Read 2 more answers

A ball is falling after rolling off a tall roof<br> The ball has what type of energy.

CaHeK987 [17]

Answer:

Linear and rotational Kinetic Energy + Gravitational potential energy

Explanation:

The ball rolls off a tall roof and starts falling.

Let us first consider the potential energy or more specifically gravitational potential energy ( $mgh$ ; $m$ = mass of the ball, $g$ = acceleration due to gravity, $h$ = height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.

Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy ( $\frac{1}{2}I\omega^2$ ; $I$ =moment of inertia of the ball & $\omega$ = angular velocity).