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nevsk [136]
3 years ago
13

a skydiver jumps out of a plane. describe how gravitational potential energy changes as the skydiver falls. describe how the sky

divers kinetic energy changes when the parachute opens
Physics
1 answer:
Oliga [24]3 years ago
4 0

gravitational potential is directly proportional to the height of the object relative to a reference line and is given as

PE = mgh                        

where m = mass of object , g = acceleration due to                    

gravity   and  h = height of the object above the reference line .


as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.

as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant

KE + PE = Total energy

so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.


when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens

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3 years ago
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3 0
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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

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Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

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2 years ago
A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp
Arte-miy333 [17]

Answer:

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E and F

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√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

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Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0
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