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Nikolay [14]
3 years ago
12

Does the initial velocity of an object have anything to do with its acceleration? For example, compared to dropping an object, i

f you throw it downward would the acceleration be different after you released it?
Physics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

Explanation:

This is an excellent question to get an answer for. It teaches you much about the nature of physics.

The answer is no.

The distance will be quite different. The time might be different in getting to the distance.  But the acceleration will be the same in either case.

How do you know? Look at one of the formulas, say

d = vi * t + 1/2*a * t^2

What does vi do? vi will alter both t and d. if vi = 0 then both d and/or t will be found. But what will "a" do? Is there anything else acting in the up or down line of action? You should answer no.

If vi is not zero, t will be less and d will take less time to get where it is going.

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One gallon of paint (volume = 3.79 X 10-???? m3) covers an area of 25.0 m2. What i the thickne s of the fresh paint on the wall?
Drupady [299]

Answer:

Explanation:

Given

Volume of paint is V=3.79\times 10^{-3}\ m^3

Area of cover A=25\ m^2

Suppose paint to be a rectangular box with thickness t and volume V

therefore we can write as

V=A\times t

t=\frac{V}{A}

t=\frac{3.79\times 10^{-3}}{25}

t=1.516\times 10^{-3}\ m

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Light is traveling through the different media shown. In which medium does light travel fastest?
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3 years ago
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Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

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72 km/hr. ...jshdhdhddjidididdiididudd

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