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
D), increases. The object absorbs light energy which in turn (energy is energy) usually involves absorbing heat as well.
Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>
Using moving water to produce electricity is an example of changing one form of energy into another form of energy.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The angle between shuttle's velocity and the Earth's field is 
Explanation:
From the question we are told that
The length of eire let out is 
The emf generated is 
The earth magnetic field is 
The speed of the shuttle and tether is 
The emf generated is mathematically represented as
making 
the subject of the formula
![\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20%20sin%20%5E%7B-1%7D%5B%20%5Cfrac%7B%5Cepsilon%7D%7BL%20%20%2A%20B%20%20%2Av%7D%20%5D)
substituting values
![\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20%20sin%20%5E%7B-1%7D%5B%20%5Cfrac%7B40%7D%7B250%20%20%2A%20%285%2A10%5E%7B-5%7D%29%20%20%2A%287.80%20%2A10%5E%7B3%7D%29%7D%20%5D)
