Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
Answer:The frequency of the echo is slightly decreased
Explanation:
Given
speed of boat 
cliff is
away
when boat is still , suppose t is the time taken by the echo to reach observer on the boat
But as soon as boat starts moving the distance between cliff and boat decreasing and time for echo to reach observer also decreases
and we know 
therefore frequency of the echo slightly decreased.
Answer:
1585.67N
Explanation:
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Answer:
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.