Answer:
Explanation:
delta V = v * alpha * delta T
= V * 0.00053 * (92.2 - 55.0)
= 0.019716 V
percentage that the owner
= [delta V / V] * 100
= [0.019716 V / V] * 100
= 1.9716 %
Answer: 909 m/s
Explanation:
Given
Mass of the bullet, m1 = 0.05 kg
Mass of the wooden block, m2 = 5 kg
Final velocities of the block and bullet, v = 9 m/s
Initial velocity of the bullet v1 = ? m/s
From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve
m1v1 = (m1 + m2) v, on substituting, we have
0.05 * v1 = (0.05 + 5) * 9
0.05 * v1 = 5.05 * 9
0.05 * v1 = 45.45
v1 = 45.45 / 0.05
v1 = 909 m/s
Thus, the original velocity of the bullet was 909 m/s
Answer:
Higher, Windward side, Condenses
Explanation:
The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.
Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.
Answer:
The radius of the wetter area expands at a rate of 
milimeters per second when radius is 150 milimeters.
Explanation:
From Geometry we remember that area of a circle is described by this expression:
(Eq. 1)
Where:
- Radius of the circle, measured in milimeters.
- Area of the circle, measured in square milimeters.
Then, the rate of change of the area in time is derived by concept of rate of change, that is:
(Eq. 2)
Where:
- Rate of change of radius in time, measured in milimeters per second.
- Rate of change of area in time, measured in square milimeters per second.
Now the rate of change of radius in time is cleared within equation above:
If we know that 
and 
, then the rate of change of radius in time is:
![\frac{dr}{dt} = \left[\frac{1}{2\pi\cdot (150\,m)} \right] \cdot \left(4\,\frac{mm^{2}}{s} \right)](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%20%3D%20%5Cleft%5B%5Cfrac%7B1%7D%7B2%5Cpi%5Ccdot%20%28150%5C%2Cm%29%7D%20%5Cright%5D%20%5Ccdot%20%5Cleft%284%5C%2C%5Cfrac%7Bmm%5E%7B2%7D%7D%7Bs%7D%20%5Cright%29)
The radius of the wetter area expands at a rate of milimeters per second when radius is 150 milimeters.
The answer is <span>D) huge masses of magma pushing sedimentary rock upward</span>
