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satela [25.4K]
3 years ago
10

Is the expression "The bigger they are, the harder they fall" a generally true statement since, in the absence of air resistance

, all objects fall with the same acceleration due to gravity? Pick the most correct answer.
Yes, because as a heavier man falls, he comes closer to the center of the earth and he falls faster.
No, since the acceleration is the same, each man will be affected the same by it as he falls.
Yes, since F = ma, therefore "a" will be the same in each case, but the mass of each will be different.
Yes, since F = ma, the heavier an object is, the more it will accelerate and therefore fall with more force.
Physics
2 answers:
lions [1.4K]3 years ago
6 0
No, because in oxygen depraved rooms, if you drop a feather and a bowling ball at the same height and time, they will fall at the same speed and have the same amount of impact.
vaieri [72.5K]3 years ago
4 0

Answer:

No, since the acceleration is the same, each man will be affected the same by it as he falls.

Explanation:

In the problem, we are told that air resistance is gone. The reason a heavier man would fall harder is because the heavier a man, the less the air resistance would affect him. The air resistance is arguably the only factor that applies in what makes heavier men fall harder. So, if the aire resistance is gone and both mean are falling at exactly the same speed, then they will hit the ground equally as hard, making the correct answer B.

Hope this helps and makes sense.

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power of a crane is 25000 watt calculate the time required by it to lift a load of 6000 kg up tp the height of 20 m​
saul85 [17]
Solution:

We have,

Power [P] = 25000 Watt

Mass [m] = 6000 kg

Height [h] = 20 metres

Time [t] = ?

Now,

P = W/t = F x d/t = mxgx h/t

Or, 25000 = 6000 x 10 x 20/25000 [.......g = 10

m/s^2]

Or, t = 6000 x 10 x 20/25000

Or, t = 1200/25

Therefore, t = 48 second

Hence, the required time for the crane to lift the load is 48 seconds.
8 0
3 years ago
To determine a waves' frequency, you must know the
kykrilka [37]
B. number of oscillations in a given period of time.
7 0
3 years ago
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A simple pendulum has a period of 2.5 s. What is its period if its length is increased by a factor of four?
Svetach [21]

Answer:

Its period if its length is increased by a factor of four is 5 s.

Explanation:

The period of a simple pendulum is given by;

T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} } \\\\ \frac{T^2}{4\pi^2} = \frac{l}{g}\\\\\frac{T^2}{l} = \frac{4\pi^2}{g} \\\\let \ \frac{4\pi^2}{g}  \ be \ constant \\\\\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\

Given;

initial period, T₁ = 2.5

initial length, = L₁

new length, L₂ = 4L₁

the new period, T₂ = ?

\frac{T_1^2}{l_1}  = \frac{T_2^2}{l_2} \\\\T_2^2 = \frac{T_1^2 l_2}{l_1} \\\\T_2 = \sqrt{\frac{T_1^2 l_2}{l_1}} \\\\  T_2 = \sqrt{\frac{(2.5)^2 \ \times \ 4l_1}{l_1}}\\\\  T_2 =\sqrt{(2.5)^2 \ \times \ 4}\\\\T_2 = \sqrt{25} \\\\T_2 = 5\ s

Therefore, its period if its length is increased by a factor of four is 5 s.

5 0
2 years ago
You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω
Aloiza [94]

Answer:

\tau=0.23\;second

Explanation:

Given,

C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\

Time constant

\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second

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3 years ago
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Answer:

que te importa mmg

Explanation:

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