Answer:
Action: Gravity pulls on the ball.
Reaction: The ball falls to the ground.
Explanation:
The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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Answer:
2856.96 J
0
0
6.78822 m/s
Explanation:
= Initial velocity = 9.6 m/s
g = Acceleration due to gravity = 9.81 m/s²
h = Height
The athlete only interacts with the gravitational potential energy. Air resistance is neglected.
At height y = 0
Kinetic energy
At height y = 0 the potential energy is 0 as
At maximum height her velocity becomes 0 so the kinetic energy becomes zero.
As the the potential and kinetic energy are conserved
The general equation
Half of maximum height
The velocity of the athlete at half the maximum height is 6.78822 m/s
Since each student emits 100 W, so 170 students will emit:
total heat = 100 W * 170 = 17,000 W
Convert minutes to seconds:
time = 50 min * (60 s / min) = 3000 s
The energy is therefore:
E = 17,000 W * 3000 s
<span>E = 51 x 10^6 J = 51 MJ</span>
Answer:
0.75 m³/s
Explanation:
Applying,
Q = vA.................... Equation 1
Where Q = flow rate of the water, v = velocity of the water, A = area of the pipe.
From the question,
Given: v = 2.5 m/s, A = 0.3 m²
Substitute these values into equation 1
Q = 2.5(0.3)
Q = 0.75 m³/s
Hence the flow rate of water in the pipe is 0.75 m³/s
