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defon
3 years ago
12

An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?

Physics
1 answer:
Leya [2.2K]3 years ago
3 0
41.2 = h-1/2g(t-1)^2 
<span> {-h = -1/2gt^2-1/2g+g*t-41.2 
</span><span> {h = 1/2gt^2 
</span><span> summing them up 
</span><span> 0 = -1/2g+g*t-41.2 
</span><span> 41.2 +4.9 = g*t 
</span><span> t = 46.1/9.8 = 4.70 sec 
</span><span> h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m </span>
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During the first phase of acceleration we have:
v o = 4 m/s;  t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?;  v o = 13 m/s; a = 1.125 m/s² ;  t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
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3 years ago
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Speed = distance / time

           = d (in meters m) / t (in seconds s) = m/s

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1 year ago
What is the condition for an object experiencing free fall?
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Answer:

The body acts under the influence of gravity.

Explanation:

An object experiencing free fall is acting under the influence of gravitational force and the acceleration due gravity is positive for any falling object. The body is able to fall freely due to the effect of gravity on it. This gravity effect causes the body to get attracted to the earth's gravitational surface due to gravitational pull exerted on the body.

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3 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
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Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

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If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

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So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

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