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ycow [4]
3 years ago
14

How much work is required to lift a 2 kg mass to a height of 10 meters?

Physics
1 answer:
scoray [572]3 years ago
6 0
Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
kg {m}^{2}   {s}^{ - 2}
This is the unit of the energy
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Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197
Leno4ka [110]

(a) 3.58 km 45° south of east

The total displacement is given by:

d=vt

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

t=169 min \cdot 60 s/min = 10140 s

Therefore, the magnitude of the displacement is

d=(3.53)(10140)=35794 m = 3.58 km

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

v_a = 2.00 m/s

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v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed (v_a). Since these two vectors are in opposite direction, we have

v= v'-v_a

Therefore we find v', Allen's velocity relative to the air:

v'=v+v_a = 3.53 + 2.00 = 5.53 m/s

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

d'=v't

and substituting,

d'=(5.53)(10140)=56074 m = 56.1 km

And the direction is the same as that of the velocity, therefore will be 90° south of east.

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Answer:

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