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2 years ago

How much work is done by a crane that lowers 1000 N of material a distance of 150 m?

Physics

1 answer:

marusya05 [52]2 years ago

3 0

Answer:

Force= 1000 Newtons. Distance= 150 metres. Work= 150000 Joules. So, the work done by crane that lowers 1000 Newton's of a material a distance of 150 meters is 150000 Joules.

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3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

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a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

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3 years ago

How much work is done by a crane that lowers 1000 N of material a distance of 150 m?​

How much work is done by a crane that lowers 1000 N of material a distance of 150 m?