Answer:
47 m/s
Explanation:
golf club mass, mc = 180 g
golf ball mass, mb = 46 g
initial golf club speed, vc1 = 47 m/s
final golf club speed, vc2 = 35 m/s
initial golf ball speed, vb1 = 0 m/s
final golf ball speed, vb2 = ? m/s
The total momentum is conserved, then:
mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2
Replacing with data and solving (dimension are omitted):
180*47 + 46*0 = 180*35 + 46*vb2
vb2 = (180*47 - 180*35)/46
vb2 = 47 m/s
Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²
Explanation:
The best and most correct answers among the choices provided by your question are he second and third choices.
<span>The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.</span>
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Answer:
1.69515 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m
Time = Distance / Speed

The reaction time cannot be more than 1.69515 seconds
1)
Answer:
Part 1)
H = 30.6 m
Part 2)
t = 2.5 s
Part 3)
t = 2.5 s
Part 4)

Explanation:
Part 1)
initial speed of the ball upwards

so maximum height of the ball is given by



Part 2)
As we know that final speed will be zero at maximum height
so we will have



Part 3)
Since the time of ascent of ball is same as time of decent of the ball
so here ball will same time to hit the ground back
so here it is given as
t = 2.5 s
Part 4)
since the acceleration due to earth will be same during its return path as well as the time of the motion is also same
so here its final speed will be same as that of initial speed
so we have

2)
Answer:
a = 9.76 m/s/s
Explanation:
As we know that the object is released from rest
so the displacement of the object in vertical direction is given as



3)
Answer:
v = 29.7 m/s
Explanation:
acceleration of the rocket is given as

time taken by the rocket
t = 0.33 min
final speed of the rocket is given as



4)
Answer:
Part 1)
y = 25.95 m
Part 2)
d = 6.72 m
Explanation:
Part 1)
As it took t = 2.3 s to hit the water surface
so here we will have



Part 2)
Distance traveled by it in horizontal direction is given as


