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MrRa [10]
3 years ago
9

The mass of a high speed train is 4.5×105 kg, and it is traveling forward at a velocity of 8.3×101 m/s. Given that momentum equa

ls mass times velocity, determine the values of m and n when the momentum of the train (in kg⋅m/s) is written in scientific notation.
Physics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

The value of m = 3.735 and the value of n = 7.

Explanation:

The equation for the momentum of the train is,

                                          P = mv  

Here, m is the mass of the train and v is the speed of the train.

Substitute 4.5 \times {10^5}{\rm{ kg}} , 8.3 \times {10^1}{\rm{ m/s}} for m and v respectively in above equation.

\begin{array}{c}\\P = \left( {4.5 \times {{10}^5}{\rm{ kg}}} \right)\left( {8.3 \times {{10}^1}{\rm{ m/s}}} \right)\\\\ = 37.35 \times {10^6}{\rm{ kg}} \cdot {\rm{m/s}}\left( {\frac{{{{10}^1}{\rm{kg}} \cdot {\rm{m/s}}}}{{10\,{\rm{kg}} \cdot {\rm{m/s}}}}} \right)\\\\ = 3.735 \times {10^7}{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}  

According to the scientific notation, here the value of m is 3.735 and the value of n is 7 in the final answer of the momentum.

                        The value of m = 3.735 and the value of n = 7.

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2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

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Therefore the potential at the center of the sphere is:

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3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

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- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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