A pitcher releases a fastball that moves toward home plate.

1 answer:

5 0

The two additional forces that act on the ball as it travels between the pitcher and the home plate are air resistance and gravity.

<h3>What are the forces that affect object in motion;</h3>

Air resistance: this is the force that oppose the motion of an object in air due to frictional force
Gravity: this is the force due to weight of the object and acts downwards.

The two additional forces that act on the ball as it travels between the pitcher and the home plate include:

Air resistance and
Gravitational force

<h3>How the forces affect the motion of the ball</h3>

Air resistance oppose the motion of the ball as it travels in air.
Gravity is the force due to weight of the ball and acts downwards.

Learn more about forces on object in motion here: brainly.com/question/10454047

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You place ammonium nitrate crystals in water and stirred. As you do so, the container becomes cold to the touch. This an example

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Answer: Option B

<u>Explanation:</u>

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If the heat is transferred from the surrounding to the system , then it is an endothermic reaction. And in those cases, the sample holder will be becoming colder. This is because the heat from the surrounding that is the container will be utilized to complete the reaction.

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3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$