Answer:
The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"
Explanation:
In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;
<em>1) v=u+at</em>
<em>2)v^2=u^2+2as</em>
<em>3)s=ut+(1/2)(at^2)</em>
the variables are explained below;
u= initial velocity of the body
a=acceleration/Speed of the body
t= time taken by the body while travelling
s= displacement of the body.
Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)
To change from mass to weight is Fw = 30 kg * 9.8 m/s^2 = 294 N. To change from weight to mass divide by gravity (9.8 m/s^2).
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,
.
Here 
are initial mass and velocity. 
are final mass and velocity. Here 
and 
.
The velocity of the truck be after its cargo is taken off is
