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Mazyrski [523]
3 years ago
8

How do you think air resistance affects measured values of g? If you used a ping pong ball, for example, how would this affect t

he fall-time? Would you expect the ping pong ball and the steel ball to land at the same time if they were both dropped from a height of ϭ.ϱm? What would happen if you dropped both balls from a height of ϮϬm?
Physics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer:

A) Air resistance acts in a direction opposite the the fall of an object reducing it by doing work against the weight of the object due to gravity.

B) using a ping pong ball, the time of fall will be greatly reduced since it has little weight (its mass x acceleration due to gravity) against the air resistance. The net downward force of the weight and the air resistance will be small.

C) No, I wouldn't expect them to fall at the same time. The steel ball will have more weight compared to the ping pong ball and hence it will have a larger net force downwards.

D) If they are both released from a 6 m height, the steel ball will fall to the ground first since it has a larger net force downwards.

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an object with a mass of 6 kilograms accelerates 4.0 MS to the second when an unknown force is applied to it what is the amount
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F=ma \\ F=6*4 \\ \boxed {F=24N}
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An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the accelerati
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Answer:

A) \frac{g}{b}(1-e^{-bt})

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since \frac{dv}{dt}= g - bv = b( \frac{g}{b} - v) ⇒ \frac{dv}{ \frac{g}{b} - v}= bdt

So take the integral of both side.

- ln (\frac{g}{b} - v) = bt + C

Since for t=0, v = 0 ⇒ C =- ln (\frac{g}{b})

v = \frac{g}{b} + e^{-bt-ln(\frac{g}{b})} = \frac{g}{b}- \frac{g}{b}e^{-bt} = \frac{g}{b}(1-e^{-bt})

5 0
3 years ago
Does the following decrease or increase normal force? Pulling up on the object.
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Answer:

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6 0
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2 years ago
A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t
Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

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m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}at^2

a=\dfrac{2s}{t^2}

a=\dfrac{2\times 12\ m}{(1.2\ s)^2}

a = 16.67 m/s²

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q = 0.0000249 C

or

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