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2 years ago

Which mass is undergoing to the greatest amount of acceleration ??

Physics

1 answer:

lina2011 [118]2 years ago

4 0

Answer:

Option (3)

Explanation:

Formula used to calculate acceleration is,

F = ma

Where F = force exerted on a mass

m = mass

a = acceleration due to force exerted on the mass

Option (1),

When F = 100 N and m = 100 kg

100 = 100a

a = 1 m per sec²

Option (2)

For F = 1 N and m = 100 kg

1 = 100a

a = $\frac{1}{100}$

a = 0.01 m per sec²

Option (3)

For F = 100 N and m = 1 kg

100 = 1(a)

a = 100 m per sec²

Option (4)

For F = 1 N and m = 1 kg

1 = 1(a)

a = 1 m per sec²

Therefore. acceleration in Option (3) is the maximum.

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Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto

Luda [366]

As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.

4 0

2 years ago

Read 2 more answers

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$