Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
The work is path independent since we have a conservative force.
Thus
Answer (1)
Answer:
= 201.53 meters
Explanation:
A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?
Use the formula d = 
where d is the distance, t is the time and "a" is the acceleration.

Correct order, from lowest potential energy to highest potential energy:
E - C - D - B - A
Explanation:
The gravitational potential energy of the car is given by:

where
m is the car's mass
g is the gravitational acceleration
h is the height of the car relative to the ground
In the formula, we see that m and g are constant, so the potential energy of the car depends only on its height above the ground, h. The higher the car from the ground, the larger its potential energy. Therefore, the position with least potential energy will be E, since the height is the minimum. Then, C will have more potential energy, because the car is at higher position, and so on: the position with greatest potential energy is A, because the height of the car is maximum.
Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by

where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by

now, wavelength calculation
n = 1

λ₁ = 4 m
n = 2

λ₂ = 2 m
n =3

λ₃ = 1.333 m
now, frequency calculation
n = 1


f₁ = 12.5 Hz
n = 2


f₂= 25 Hz
n = 3


f₃ = 37.5 Hz