Physics, 17.07.2020 06:01 ellamai16

How does gravity, inertia and friction play a role in throwing a ball

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Answer from: moongirl4791

The role of gravity, inertia and friction play different roles when a ball s thrown such as the following;

The thrown ball (which was initially at rest) is given its inertia by the force delivered to it (the ball) by the hand of the thrower of the ball

Force applied to the hand by the thrower of the ball has to have a balancing equal and opposite force (reaction force) for the person to remain in position (not skidding or falling over)

The reaction force mentioned above is transmitted from the hands through the body to the legs which rests on the ground

The leg pushes in the opposite direction in which the ball is thrown but due to combination of the gravity pull which is the weight which acts downward and the friction resistance which acts sideways and increases proportionally with the weight of the thrower due to gravity, the reaction force in the leg is not able to overcome the frictional resistance force of the leg and the thrower remains upright while the throwing of the ball is successful

Gravity will then act on the ball to limit the distance traveled such that the ball will not wander away and could be directed at desired target

Explanation:

Answer from: Quest

He can play two games. 13/4=3.253.25*2=6.5

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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