Answer:
The algebraic equation is:
Explanation:
Given information:
mb = book's mass
vb = tangential speed
R = radius of the path
Question: Derive an algebraic equation for the vertical force, Fv = ?
To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:
The variables were defined above and g is the gravity.
Answer:
Explanation:
Mass of the cable car, m = 5800 kg
It goes 260 m up a hill, along a slope of 
Therefore vertical elevation of the car = 
Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = 
). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.
Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.
Hence, total change in energy = mgh = 
where, g = acceleration due to gravity
h = height/vertical elevation
Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
Answer:
Emitted power will be equal to 
Explanation:
It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m
Threshold of human hearing 
We have to find the emitted power
Emitted power is equal to 
So emitted power will be equal to
Answer:
The rate of change of the height is - 4 ft/s
Solution:
As per the question:
Height of the person, y = 5 ft
The rate at which the person walks away, 
Distance of the spotlight from the wall, x = 40 ft
Now,
To calculate the rate of change in the height, 
of the person when, x = 10 m:
From fig 1.
![\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}](https://tex.z-dn.net/?f=%5CDelta%20ABC%5B%5Ctex%5D%20%E2%89%88%20%5Btex%5D%5CDelta%20PQC%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3EThus%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BBC%7D%7BAB%7D%20%3D%20%5Cfrac%7BPQ%7D%7BQC%7D)
xy = 200 (1)
Differentiating the above eqn w.r.t time t:
Thus
(2)
From eqn (1):
When x = 10 ft
10y = 200
y = 20 ft
Using eqn (2):
