Physics, 08.04.2020 02:12 kyleeb00

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms.

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A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30∘ above horizontal with a speed of 52.0 m/s. The ball and bat are in contact for 1.85 ms, find the horizontal and vertical components of the average force on the ball.

The average horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.

Explanation:

Given:

Mass of the baseball, = 0.145 kg

Velocity before the impact, v_1 = 40 m/s

Velocity after the impact, v_2 = 52 m/s

Angle of deflection, $\theta$ = 30 deg

Time of contact, $\triangle t$ = 1.85 ms

Conversion of milliseconds to seconds.

⇒ $\triangle t = 1.85\times 10^-^3\ sec$

Let the horizontal and vertical component of 'v1' and 'v2' be:

⇒ v_1x,v_1y,v_2x,v_2y

⇒ $v_1x=40\ ms^-^1$ and $v_2y= 0\ ms^-^1$

⇒ Taking the vector components of 'v2'

⇒ $v_2x=52cos(30) = 45.03\ ms^-^1$ and $v_2y=52sin(30)=40.97\ ms^-^1$

We have to find the impulse in x-y, direction.

Momentum with 'v1'

⇒ P_1_x=mv_1_x

⇒ P_1_x=0.145(40)

⇒ $P_1_x=5.8\ kg.m.s^-^1$ and P_1_y=0

Momentum with 'v2'

⇒ P_2_x=mv_2_x ⇒ P_2_y=mv_2_y

⇒ P_2_x=0.145(45.03) ⇒ P_2_y=0.145(40.97)

⇒ $P_2_x=6.52\ kg.m.s^-^1$ ⇒ $P_2_y=5.96\ kg.m.s^-^1$

Now the difference in terms of impulse (J).

Horizontally Vertically

⇒ J_x=P_2_x-P_1_x ⇒ J_y=P_2_y-P_1_y

⇒ J_x=6.52-5.8 ⇒ J_y=5.96-0

⇒ $J_x=0.72\ kg.m.s^-^1$ ⇒ $J_y=5.96\ kg.m.s^-^1$

Now we know that Impulse is the product of force and time.

So,

Horizontal force : Vertical force:

⇒ $(F_x)_a_v_g=\frac{J_x}{\triangle t}$ ⇒ $(F_y)_a_v_g=\frac{J_y}{\triangle t}$

⇒ $(F_x)_a_v_g=\frac{0.72\ kg.ms^-^1}{1.85\times 10^-^3\ s}$ ⇒ $(F_y)_a_v_g=\frac{5.96\ kg.m.s^-^1}{1.85\times 10^-^3\ s}$

⇒ $(F_x)_a_v_g=389.18\ N$ ⇒ $(F_x)_a_v_g=3221.62\ N$

So the horizontal force is 389.18 N and vertical component of the average force is 3221.62 N.

A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the rig

Answer from: Quest

100 pounds is 100 pounds no matter the object or item being weighed

Answer from: Quest

explanation:

to push or pull an object in a certain direction is known as force. a force in moving an object from one place to another.

force depends on the mass of an object. so, more is the mass of an object more will be the force required to move the object.

thus, we can conclude that the use of force to move an object is push or pull.

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