Answer:
The Role of Heat Transfer Methods in the Distribution of Earth's Energy
Explanation:
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.
So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.
Force on the student = 60 kg * 2.0 m/s^2 = 120 N.
=> horizontal force exerted by the student on the boat = 120 N
Answer: option d. 120 N. toward the back of the boat.
Of course it is toward the back because that is where the student jumped from..
Lets write again formula for determening Ann's performance.
P = 70.443 + 4,885*t
where t is in hours. This is equation with P(t) which means that P only depends on variable t. If we express t=2.6 in formula we will find her expected performance.
P = 83.144
Now, since it says that she scored 16 points less than expected we need to find value of P-16
P - 16 = 67.144
After round we get that the answer is 67
Explanation:
The electric field at a distance r from the charged particle is given by :

k is electrostatic constant
if r = 2 m, electric field is given by :

If r = 1 m, electric field is given by :

Dividing equation (1) and (2) we get :

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
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):
