The task is to show that the right side of the equation has units of [Time], just like the left side has.
The right side of the equation is . . . 2 π √(L/G) .
We can completely ignore the 2π since it has no units at all, so it has no effect on the units of the right side of the equation. Now the task is simply to find the units of √(L/G) .
L . . . meters
G . . . meters/sec²
(L/G) = (meters) / (meters/sec²)
(L/G) = (meters) · (sec²/meters)
(L/G) = (meters · sec²) / (meters)
(L/G) = sec²
So √(L/G) = seconds = [Time]
THAT's what we were hoping to prove, and we did it !
Gs*rs^2 = gm*rm^2
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl
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Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
The electromagnetic spectrum is traditionally divided into regions of radio waves, microwaves, infrared radiation, visible light, ultraviolet rays, x rays, and gamma rays. ... Wavelength defines the distance between adjacent points of the electromagnetic wave that are in equal phase (e.g., wavecrests)