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2 days ago

What’s the force of a pitching machine on a baseball?

Physics

2 answers:

Ostrovityanka [2.2K]2 days ago

7 0

Answer:

Force, F = 9 N

Explanation:

The initial velocity of the baseball, u = 0 m/s

Final velocity of the baseball, v = 30 m/s

Time taken, t = 0.5 s

Mass of the baseball, m = 0.15 kg

The force applied by a pitching machine on a baseball can be calculated using the second law of motion. The formula is given by:

$F=ma$

$F=m\times \dfrac{v-u}{t}$

$F=0.15\times \dfrac{30}{0.5}$

F = 9 N

Thus, the pitching machine exerts a force of 9 N on the baseball. This is the required solution.

Ostrovityanka [2.2K]2 days ago

6 0

Answer:

F = 9 N

Explanation:

Acceleration of the ball can be defined as the rate at which the velocity changes

and here we commence with

$a = \frac{dv}{dt}$

noting that

$v_i = 0$

$v_f = 30 m/s$

$\Delta t = 0.5 s$

now we will find

$a = \frac{30 - 0}{0.5} = 60 m/s^2$

to determine the force, we will apply Newton's second law as follows

$F = ma$

from the prior formula we derive

$F = (0.15)(60) = 9 N$

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Keith_Richards [2256]

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

$\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m$

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

$\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s$

The person's speed is 11.56066 m/s

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27 days ago

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

Keith_Richards [2256]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.

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22 days ago

A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

kicyunya [2264]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

$U_1=k\dfrac{q_1q_2}{x_1}$

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Consequently, the work done by the electrostatic force on the moving charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Therefore, this concludes the solution.

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10 days ago

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Sav [2217]

Response: Numerous elements can be found, all situated within the same vertical column as bromine.

Explanation:

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1 month ago

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Answer:

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(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

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Based on the diagram,