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Bess [88]
4 years ago
9

A 6kg rock is thrown vertically upward from the top of a cliff that is 28m high. It was thrown with a speed of 11ms . Calculate

the greatest height of the rock as measured from the base of the cliff.
Physics
1 answer:
pickupchik [31]4 years ago
6 0

Answer:H=34.05m

Explanation: H=u²/2g+ 28m height of the cliff

u=11ms⁻¹

g=10ms⁻²

H=11²/2*10

H=121/20

H=6.05+28

H=34.05m

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Answer:

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3 years ago
When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle
LUCKY_DIMON [66]
At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.
4 0
3 years ago
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The maximum speed around a level curve is 30.0 km/h. What is the maximum speed around a curve with twice the radius?
erastovalidia [21]

Answer:

When the radius is doubled, its speed will be 42.42 km/h.

Explanation:

Given that,

The maximum speed around a level curve is 30 km/h = 8.34 m/s

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v=\sqrt{\mu rg}............(1)

If radius is doubled, r' = 2r

The maximum speed is given by :

v'=\sqrt{\mu r'g}

v'=\sqrt{\mu (2r)g}..........(2)

Dividing equation (1) and (2) we get :

\dfrac{v}{v'}=\dfrac{1}{\sqrt2}

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v'=\sqrt{2}\times 30

v' = 42.42 km/h

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4 years ago
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3 years ago
An airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highw
docker41 [41]
Using trigonometric ratios we can get the distance;
For the first car; The distance from the point on the highway below the plane 
tan = opp/adj
tan(36°) = 5150/x
0.727 = 5150/x
0.727x = 5150
x = 7088.37
For the second car we also use tangent; the distance from the point on the highway below the plane will be; 
tan(56°) = 5150/y
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7088.37 + 3473.72 = 10562.09 feet. 
= 10562.09 ft
7 0
3 years ago
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