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Ivan
3 years ago
13

1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15

0 km to station B in 2 hours. What is the average velocity of the train?
Average velocity= displacement/time

2-A car accelerates uniformly at the rate of 1.2 meters per second squared for a distance of 75 meters. If the truck started with a initial velocity of 6 meters per second, what is its velocity at the end of the 75 meters?
vf^2=vi^2 + 2as
Physics
1 answer:
kirill [66]3 years ago
8 0

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity=\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 m/s^2

Distance covered by the car,s = 75 m

Initial velocity of the car ,v_i = 6 m/s

Final velocity of thre car ,v_f=?

Using third equation of motion:

v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2

v_{f}=14.69 m/s

The final velocity of the car at the end of 75 m is 14.69 m/s

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A gold nucleus (with a radius of 6.5 fm in its rest system, containing 197 protons and neutrons of rest mass 939 MeV/c2 each) is
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Answer:

1.9982154567\times 10^{-18}\ kgm/s

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Relativistic momentum is given by

p=\dfrac{m_0v}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow p=\dfrac{939\times 10^6\times 1.6\times 10^{-19}\times 0.97\times 3\times 10^8}{\sqrt{1-\dfrac{0.97^2c^2}{c^2}}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})J/c^2\times 0.97c}{\sqrt{1-0.97^2}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})\times 0.97}{c\sqrt{1-0.97^2}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})\times 0.97}{3\times 10^8\times \sqrt{1-0.97^2}}\\\Rightarrow p=1.9982154567\times 10^{-18}\ kgm/s

The momentum is 1.9982154567\times 10^{-18}\ kgm/s

Energy in MeV

E=\dfrac{m_0c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow E=\dfrac{939}{\sqrt{1-0.97^2}}\\\Rightarrow E=3862.52987766\ MeV

Total energy is

E'=3862.52987766+7.9=3870.42987766\ MeV

The total energy is 3870.42987766 MeV

For all the nucleons

E_t=197\times 3870.42987766=762474.685899\ MeV

The energy is 762474.685899 MeV

Diameter = 2\times 6.5=13\ fm

From length contraction

D'=D\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow D'=13\sqrt{1-0.97^2}\\\Rightarrow D'=3.1603639031\ fm

The diameter would be 3.1603639031 fm

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