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Inessa05 [86]
4 years ago
11

A train travels 73 kilometers in 3 hours, and then 60 kilometers in 3 hours. what is the average speed

Physics
1 answer:
Jlenok [28]4 years ago
5 0
The answer is 22.17 km/hr. The total distance would be the sum of the first 73 kilometers and the second distance 60 kilometers, and the time the train took to cover that distance would be the sum of the first 3 hours and the second 3 hours. Since the average speed of the train is the total distance it traveled divided by the elapsed time to cover that distance, we have
     average speed = total distance/elapsed time
                              = (73km + 60km) / (3hr + 3hr)
                              = 133km / 6hr
                              = 22.17 km/hr
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Answer:

(a). The rotational inertia is 5.72\times10^{39}\ kg m^2

(b). The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

Explanation:

Given that,

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Density of neutron \rho=4.8\times10^{17}\ kg/m^3

(a). We need to calculate the rotational inertia

Using formula of rotational inertia  for sphere

I=\dfrac{2}{5}MR^2...(I)

We know that,

\rho=\dfrac{M}{V}

Put the value of volume

\rho=\dfrac{3M_{n}}{4\pi R^3}

R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}

Put the value of R in equation (I)

I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}

Put the value into the formula

I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}

I=5.72\times10^{39}\ kg m^2

The rotational inertia is 5.72\times10^{39}\ kg m^2.

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

\tau=I\times \alpha

Put the value into the formula

\tau=5.72\times10^{39}\times5.6\times10^{-5}

\tau=3.20\times10^{35}\ N-m

The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

Hence, (a). The rotational inertia is 5.72\times10^{39}\ kg m^2

(b). The magnitude of the magnetic torque is 3.20\times10^{35}\ N-m

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