V=IR however in this case we will rearrange it into R= V/I. Which leads to R=15.0/0.30. And your answer becomes 50.
Answer:
v = 88.89 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.
We must make up the following equation:

where:
F = force applied = 4000 [N]
t = time = 0.001 [s]
m = mass = 0.045 [kg]
v = velocity [m/s]
![4000*0.001=0.045*v\\v=88.89[m/s]](https://tex.z-dn.net/?f=4000%2A0.001%3D0.045%2Av%5C%5Cv%3D88.89%5Bm%2Fs%5D)
Answer:
The displacement of the air drop after 3 second is 18.27 m.
Explanation:
Mass of the rain drop = m = 
Weight of the rain drop = W
Duration of time = t = 3 seconds

Drag force on rain drop = 

Motion of the rain drop:

Net force on the rain drop , F= W - D




v = 12.18 m/s
Initial velocity of the rain drop = u = 0 (since, it is starting from rest)
v=u+at (First equation of motion)


(second equation of motion)

s = 18.27 m
The displacement of the air drop after 3 second is 18.27 m.
Answer:
a) 91 m/s
b) 111 m/s
Explanation:
v = u + at
v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s
v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s
<h3><u>Answer;</u></h3>
just before it reaches the ground
<h3><u>Explanation;</u></h3>
- Kinetic energy is the energy possessed by a body or an object in motion.
- <em><u>Kinetic energy is given by 1/2mv², where m is the mass of the object and V is the velocity of the body. Thus, kinetic energy depends on the velocity of the body if mass is kept constant.</u></em>
- <em><u>As soon as the ball leaves the racket it has more kinetic energy and zero potential energy. As it moves up its velocity decreases, and thus the kinetic energy is being converted to kinetic energy up to maximum height reached where kinetic energy will be zero since the velocity is zero.</u></em>
- <em><u>When the ball is going down the potential energy will be converted to kinetic energy up to a point just before it hits the ground, where kinetic energy is maximum since the velocity of the ball is maximum, due to gravitational acceleration.</u></em>