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castortr0y [4]
3 years ago
8

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to

1.96 m 3. what is the volume strain on the box
Physics
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02

So, the volume strain on the box is 0.02.

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
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1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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