<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:
R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km
here, the speed of the satellite is:
v = sqrt(R*9.00m/s^2) = 7730 m/s
the time it would take the satellite to complete one full rotation is:
T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h
Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
Water falls from a reservoir through a channel to a turbine. The water turns the turbines, which allows the generator to make electricity.
It is falling instead of flowing, because elavation is an important part of the hydroeletric power plant, since gravity is a thing, and there was elevation, than it would be falling and not flowing.
Answer:
w= p∆v 50000 ( 0.55-0.40) and calculate and you get it
