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Jobisdone [24]
3 years ago
10

Give two examples where we feel the presence of motion through indirect evidences

Physics
1 answer:
Elza [17]3 years ago
8 0

there are situations where the motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don’t we directly perceive the motion of the earth?

<span>An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road–side perceives the bus along with the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest</span>

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What is the value of acceleration due to gravity, g, on Earth?
qwelly [4]

Explanation:

It varies with altitude, but at sea level, it's 9.8 m/s².

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3 years ago
A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
Margarita [4]

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The answer to the question is 7200

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Provide a reason why houses are painted with dark colours in cold countries.​
fenix001 [56]

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A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

Then F = ma = 0.04 x 2500 = 100N

5 0
3 years ago
A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass o
luda_lava [24]

Answer:

Explanation:

We shall apply law of conservation of momentum to know velocity after collision . Let it be v .

total momentum before collision = total momentum after collision

15 x 1.5 - 12 x .75 = ( 15 + 12 ) v

v = .5 m /s

kinetic energy before collision

1/2 x 15 x 1.5² + 1/2 x 12 x .75²

= 16.875 + 3.375

= 20.25 J

kinetic energy after collision

= 1/2 x ( 15 + 12 ) x .5²

= 3.375 J

Loss of energy = 16.875 J

This energy appear as heat and sound energy that is produced during collision .

4 0
3 years ago
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