Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
<em><u>throwing a ball up initially has a lot of kinetic energy because it is moving upwards ( kinetic energy is energy which a body possesses by virtue of being in motion.) this all then get converted to gravitational potential energy, and for a moment it is stationary before it begins to fall again. by the time it has returned again, all the gravitational potential energy has turned back into kinetic.</u></em>
Answer:
<em> 3980.89 ohms</em>
Explanation:
The capacitive reactance is expressed as;

f is the frequency
C is the capacitance of the capacitor
Given
f = 60H
C = C1+C2 (parallel connection)
C = 15μF + 25μF
C = 40μF
C = 
Substitute into the formula:

<em>Hence the total capacitive reactance is 3980.89 ohms</em>
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Refraction. ... Diffraction. ... EM spectrum. ... Intensity. ... Transverse wave. ... Frequency. ... Compression wave.