Answer:
Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s
Explanation:
Let's start out with finding the force acting downwards because of the mass of 'The Rock':
Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N
Now the problem also states that the kinetic friction of the desk in this problem is 370 N
Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N
Now lets use F = ma to calculate for the acceleration of the desk:
787.58 = 63 x acceleration
acceleration = 12.501 m/s
Finally, we can use the motion equation:
here u = 0 m/s (since initial speed of the desk is 0)
a = 12.501 m/s
and s = 10 m
Solving this we get:
Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.
Answer:
Option (3)
Explanation:
Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.
He was the person who contributed to the heliocentric theory. This theory describes the position of the sun in the middle of the universe, and all the planets move around the sun. This theory was initially not accepted, and after about a century it was widely accepted.
This theory describes the present-day motion of the planets around the sun in the solar system. This theory replaced the geocentric theory.
Thus, the correct answer is option (3).
Acceleration occurs when velocities change.velocity changes either because the speed changes or the direction change.
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge
The charge on the inner surface is q.
We need to calculate the electric field outside the shell
Using formula of electric field
Put the value into the formula
Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is
