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3 years ago

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You (in the x' frame) are moving with v = +5 m/s with respect to the x frame. If your initial position in each frame is Om, what

are the positions in each frame after 3 s have passed? X:-15 m x': 15 m Answer 1: - 15 m Answer 2: 15 m

1 answer:

WARRIOR [948]3 years ago

3 0

Answer:

Distance is 15 meters.

Explanation:

It is given that,

Velocity with respect to the x frame, v = 5 m/s

We need to find the position in each frame after 3 seconds have passed.

Distance covered is calculated by the product of velocity and time taken. It is given by :

$d=v\times t$

$d=5\ m/s\times 3\ s$

d = 15 meters

So, after 3 seconds the position in each frame is 15 meters from origin. Hence, this is the required solution.

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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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Answer:

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Answer:

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Explanation:

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First thing you want to do is convert 250 kg to g (grams).

From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).

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Answer:

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Explanation:

divide the distance by the speed

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