Answer:
v2 = 27.3m/s
Explanation:
Assuming forward as positive.
Mass = m1 = 64kg
Let v be the common velocity of the student and the skateboard.
mass of skateboard = m2 = 5.94kg
v = 1.4m/s
Since the skateboard and the student are initially moving together at the same velocity their momentum together is
(m1 + m2)v
Let the final velocity of the student be v1 and the final velocity of the skateboard be v2
v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)
Then from conservation of momentum, momentum before is equal to momentum after.
(m1 + m2)v = m1v1 + m2v2
m2v2= (m1 + m2)v – m1v1
v2 = ( (m1 + m2)v – m1v1)/m2
v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94
v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94
v2 = 27.3m/s
I am not sure if this is the answer you are looking for but a earthquake occurs when the plates shift.
Hope this helped.
(a) Let be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude
where is the radius of the circle.
The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is
so that
Solve for :
(b) The net force equation in part (a) leads us to the relation
so that is directly proportional to the square root of . As the radius increases, the maximum linear speed will also increase, so the cord is less likely to break if we keep up the same speed.
Answer:
9.01amp
Explanation:
Power = V^2/R
Given that v = 11volts, P = 99watts
99 = 11^2/R
11×11 = 99R
121= 99R
R = 121/99
R= 1.22ohms
From ohms Law; V = IR
11volts = I × 1.22ohms
I = 11/1.23
I = 9.01 amp