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katovenus [111]
3 years ago
11

What Kinetic Energy does the sack have just before it strikes the floor?

Physics
1 answer:
murzikaleks [220]3 years ago
8 0
When an object is at any height above the ground, it has gravitational potential energy. If the object falls, it loses this gravitational potential energy and this energy is converted to kinetic energy. By this logic, we can say:

Gravitational potential energy of grain sack = Kinetic energy of grain sack

GPE = mgh
        = (mg) x h
        = 98.0 x 50.0
        = 4,900 J
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A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

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3 years ago
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The net force is 12 N to the left.
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2 years ago
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In this dark-colored forest, you started with 50% light moths and 50% dark moths. At the end of your simulation, was there a hig
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What are the Vedas?
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Answer:

B religious writing

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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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