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Tems11 [23]
3 years ago
6

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

done to lift the load 2 m, using the same effort?
A) Use an inclined plane of length 8 m.
B) Use an inclined plane of length 2 m.
C) Use an inclined plane of length 10 m.
D) Use an inclined plane of length 16 m.
Physics
2 answers:
Troyanec [42]3 years ago
6 0
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement 
so when we wanted to move a 100kg a distance of 1m 
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m 

by the same way:-
------------------------
work in = 100kg * 2m * g (m/s^2)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
insens350 [35]3 years ago
4 0
A) Use an inclined plane of length 8 m.
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An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
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Answer:

V = 0.0806 m/s

Explanation:

given data

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mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

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5 0
3 years ago
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
3 years ago
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