Answer:
The velocity of the rocket is 7.8 m/s
Explanation:
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:
kinetic energy = 0.1168 J
Explanation:
From Hooke's law, we know that ;
F = kx
k = F/x
We are given ;
Mass; m = 1.95 kg
Spring stretch; d = x = 0.0865
So, Force = mg = 1.95 × 9.81
k = 1.95 × 9.81/0.0865 = 221.15 N/m
Now, initial energy is;
E1 = mgL + ½k(x - L)²
Also, final energy; E2 = ½kx² + ½mv²
From conservation of energy, E1 = E2
Thus;
mgL + ½k(x - L)² = ½kx² + ½mv²
Making the kinetic energy ½mv² the subject, we have;
½mv² = mgL + ½k(x - L)² - ½kx²
We are given L=0.0325 m
Plugging other relevant values, we have ;
½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)
½mv² = 0.62170875 + 0.3224367 - 0.82734979375
½mv² = 0.1168 J
Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant = 
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by

The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.

The distance from the electric field is 1.71436 m