Dr. Hewitt's experiment using the bowling ball to showcase the relationship between <em>momentum, energy and speed</em> of a body in motion. In his first attempt, the ball returns and stops almost exactly at the point it was launched.
- In Dr.Hewitt's first trial with the bowling ball, the ball was launched without any additional force applied, hence, the initial energy during the launch was converted to same amount of potential energy on the ball's return. Hence, stopping at the same point where the ball was launched.
- In subsequent trials, when extra force was applied, the ball went past the <em>initial launch position</em> as the potential energy as the ball returned was higher.
Therefore, the ball stopped at the <em>position of initial launch</em> during the first trial.
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External = R
Internal = r
Volume of hemisperical = 2/3 π(R³-r³)
V= 2/3 π(9.1³ - 8.4³)
V= 336.9 cm³
Answer: 6.22 N(the units have been mentioned wrongly in the question)
Given:
mass of the ball(m)=0.42 Kg
acceleration of the ball(a)=14.8m/s^2
F=mxa
Where m is the mass of the ball.
a is the acceleration of the ball.
F is the force applied on the ball.
F=0.42 X 14.8
F= 6.26 N
Answer:
Explanation:
Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as
W = F×d
or
W = .................................. Equation 1
Where b = upper limit, a = lower limit, Fx = expression of force.
<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>
Substituting these values into equation 1
<em>W = </em>
W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ
Work = upper limit - lower limit
Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2
Substituting the values of a and b into equation 2
Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]
Work = [5.2 + 26.53 -3.29 + C] - C
Work = 28.44 J
Work done by the force = 28.44 J.