Answer:
albert einsteinssssssssssssssssssssssssss
Explanation:
Answer:
25 m/s
Explanation:
This question can be solved using equation of motion

where
v is the final velocity
u is the initial velocity
s is the distance covered while moving from initial to final velocity
a is the acceleration
_____________________________________________
Given
box moved for distance of 62.5 m
Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.
thus in this problem
a = -5.0 m/s2
V = 0 as body came to rest due to friction deceleration
u the initial velocity we have to find
the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.
so if we find speed of box, we will be able get sped of truck as well.
using equation of motion

Thus, initial speed with the truck was travelling was 25 m/s.
Answer:
The final kinetic energy is
Explanation:
From the question we are told that
The electric field is 
The charge on the object is 
The mass of the object is 
The distance moved by the object is 
The workdone on the object by the fields is mathematically represented as
![W = [qE + mg]d](https://tex.z-dn.net/?f=W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
Now this workdone is equivalent to the final kinetic energy so
![K = W = [qE + mg]d](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5BqE%20%2B%20mg%5Dd)
substituting values
![K = W = [4.5*10^{-3 } *100 + 0.68 * 9.8]* 1](https://tex.z-dn.net/?f=K%20%3D%20W%20%3D%20%20%5B4.5%2A10%5E%7B-3%20%20%7D%20%2A100%20%20%2B%200.68%20%2A%209.8%5D%2A%201)

Answer:
Granite
Explanation:
Trust me I learned this 2years ago
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.