Answer:
The answer to your question is 784.8 J. None of your answer, did you forget some information?
Explanation:
Data
mass = 20 kg
distance = 4 m
work = ?
Formula
Work = force x distance
Force = mass x gravity
Process
1.- Calculate the weight of the block
Weight = 20 x 9.81
Weight = 196.2 N
2.- Calculate the work done
Work = 196.2 x 4
Work = 784.8 J
Answer:
a)
b)
c)
Explanation:
We use the definition of a electric field produced by a point charge:

<u>a)Electric Field due to the alpha particle:</u>

<u>b)Electric Field due to electron:</u>

<u>c)Electric Force on the alpha particle, on the electron:</u>
The alpha particle and electron feel the same force but with opposite direction:

Answer:
a) The car was moving at a speed of 
b) The negative sign of
denotes that the observer is coming towards the police car which is the source of the sound.
c) 
Explanation:
Given:
- original frequency of the source,

- speed of the source,

- velocity of the obstacle car be,

- speed of sound,

- observed frequency,

<u>Using the equation from the Doppler's effect:</u>



a)
The car was moving at a speed of 
b)
The negative sign of
denotes that the observer is coming towards the police car which is the source of the sound.
c)
Now when, 
Then, 
Using the Doppler's eq.:


Answer:
The value is 
Explanation:
From the question we are told that
The initial pressure is
The initial temperature is ![T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%2050%20%5C%20F%20%3D%20%2850%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D%20283%20%20%5C%20%20K)
The final temperature is ![T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20320%20%5C%20F%20%3D%20%28320%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D433%20%20%5C%20%20K)
Generally the equation for adiabatic process is mathematically represented as

=> 
Generally for a monoatomic gas 
So
![14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }](https://tex.z-dn.net/?f=14%20%2A%20283%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D%20%3DP_2%20%2A%20433%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D)
=> 
=> 
Answer:
The frictional force is 
Explanation:
From the question we are told that
The coefficient of kinetic force is μk = 0.35
The normal force felt by the puck is 
Generally the frictional force that acts on the puck is mathematically represented as

=> 
=> 