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Vilka [71]
3 years ago
9

Your initial speed is 12 m/s and your final speed is 13 m/s. If this change of speed took place over a distance of 156 m how lon

g did it take.
Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0

Answer:

The answer to your question is: 12.48 s

Explanation:

Data:

initial speed (is) = 12 m/s

final speed (fs) = 13 m/s

distance (d) = 156 m

time (t) = ?

Formula

d = ((fs + is))/2 t        we clear time from this equation    t = 2d / (fs + is)

Substitution             t = 2(156) / (13 + 12)

Process                    t = 312 / 25

                                 t = 12.48 s

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a brick is suspended above the ground at a height of 6.6 meters. it has a mass of 5.3 kg. what is the potential energy of the br
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Answer:

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(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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