Answer:
The correct option is;
A. The potential energy between both like charges and like poles increases as they move closer together
Explanation:
Here we have that when we move the like poles of two bar magnets close to each other, there is an increased resistance in the continuing motion, therefore for each extra gap closer achieved, there is an increase in potential energy
Similarly, when two like charges are brought closer together, the potential energy, or the energy available to push the two like charges apart increases charge as the as the charges are brought closer together
Therefore, the correct option is the potential energy between both like charges and like poles increases as they move closer together.
We know that the source of light in the universe is the Sun. Hence, the light we see as moonlight travels from the Sun's surface, to the moon, then to Earth. So, before being able to solve this problem, we have to know the distance between the Sun and the moon, and the distance between the moon and Earth. In literature, these values are 3.8×10⁵ km (Sun to moon) and 384,400 km (moon to Earth). Knowing that the speed of light is 300,000 km per second, then the total time would be
Time = distance/speed
Time = (3.8×10⁵ km + 384,400 km)/300,000 km/s
Time = 2.548 seconds
Thus, it only takes 2.548 for the light from the Sun to reach to the Earth as perceived to be what we call moonlight.
Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so

=> 

integrating both sides

Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have

![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
