Natural gas is very abundant in the United States. Compared to other fossil fuels, which describes the waste products of the com
2 answers:
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Answer:
a. b. D = v₀²/2g
Explanation:
Here is the complete question
A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance
Find the time at which the two balls collide.
Express your answer in terms of the variables D ,v₀ , and appropriate constants..
t = ?!
Part B
Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.
Express your answer in terms of the variables v₀ and g .
D =?!
Solution
The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.
The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.
The two balls collide when their vertical distances are equal. That is H₁ = H₂
So, D - v₀t + gt²/2 = v₀t - gt²/2
Collecting like terms
D + gt²/2 + gt²/2 = v₀t + v₀t
D +gt² = 2v₀t
gt² - 2v₀t + D = 0.
Using the quadratic formula,
B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.
0 = v₀² - 2gs
s = v₀²/2g.
Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,
0 = v₀ - gt₀
t₀ = v₀/g
Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.
So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g
<h2>Answer:
</h2>
The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting
the radius of the orbit (measured from the center of the planet to the satellite).
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>
In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.
Answer:
a) s
b) 3.41 mm
Explanation:
a)
We take the speed of light, c = m/s and the refractive index of glass as 1.517.
Speed = distance/time
Time = distance/speed
Refractive index, n = speed of light in vacuum / speed of light in medium
b)
We take the refractive index of water as 1.333.
Speed in water = speed in vacuum / refractive index of water
Distance = speed * time
d = 3.41 mm