The energy of the ski lift at the base is kinetic energy:

where m is the mass of the ski lift+the people carried, and

is velocity at the base.
As long as the ski lift goes upward, its velocity decreases and its kinetic energy converts into potential energy. Eventually, when it reaches the top, its final velocity is v=0, so no kinetic energy is left and it has all converted into gravitational potential energy, which is

where

and h is the height at the top of the hill.
So, since the total energy must conserve, we have

and so

from which we find the height:
Explanation:
It is given that,
Kinetic energy of the electron, 
Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.
The magnetic field in north direction, 
The magnetic field in west direction, 
The magnetic field in vertical direction, 
Magnetic field, 
Firstly calculating the velocity of the electron using the kinetic energy formulas as :



(as it is moving from west to east)
The force acting on the charged particle in the magnetic field is given by :


Since, 
And, 
![F=1.6\times 10^{-19}\times [1178 k-2864.20j]](https://tex.z-dn.net/?f=F%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20%5B1178%20k-2864.20j%5D)


(b) Let a is the acceleration of the electron. It can be calculated as :



Hence, this is the required solution.
Answer:
a = 52s²
Explanation:
<u>How to find acceleration</u>
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
<u>Solve</u>
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
Answer:
pi / 2 radians / s
Explanation:
One revolution = 2 pi Radians in 4 seconds
2 pi / 4 = pi/2 radians / s
The answer to the correct number of significant figures is 6.774.
<h3>What is quotient?</h3>
When a number(big) divided smaller number, the answer obtained greater than zero is called a quotient.
Divide 143.6 ÷ 21.2
143.6/21.2 = 1436/212
=6.77358
The quotient is rounded to three significant figures after decimal
143.6 ÷ 21.2 = 6.774
Thus, the answer to the correct number of significant figures is 6.774
Learn more about quotient
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