Answer:
ΔP = 14.5 Ns
I = 14.5 Ns
ΔF = 5.8 x 10³ N = 5.8 KN
Explanation:
The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:
ΔP = mv₂ - mv₁
ΔP = m(v₂ - v₁)
where,
ΔP = Change in Momentum = ?
m = mass of ball = 0.145 kg
v₂ = velocity of batted ball = 55.5 m/s
v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)
Therefore,
ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)
<u>ΔP = 14.5 Ns</u>
The impulse applied to a body is equal to the change in its momentum. Therefore,
Impulse = I = ΔP
<u>I = 14.5 Ns</u>
the average force can be found as:
I = ΔF*t
ΔF = I/t
where,
ΔF = Average Force = ?
t = time of contact = 2.5 ms = 2.5 x 10⁻³ s
Therefore,
ΔF = 14.5 N.s/(2.5 x 10⁻³ s)
<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>
Answer:
![\tau = 7.63 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%207.63%20Nm)
Explanation:
As we know that moment of force is given as
![\tau = \vec r \times \vec F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cvec%20r%20%5Ctimes%20%5Cvec%20F)
now we have
![\vec r = 1.2 m](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%201.2%20m)
![\vec F = 14 N](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%2014%20N)
now from above formula we have
![\tau = r F sin\theta](https://tex.z-dn.net/?f=%5Ctau%20%3D%20r%20F%20sin%5Ctheta)
here we know that
![\theta = 27 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2027%20degree)
so we have
![\tau = (1.2)(14) sin27](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%281.2%29%2814%29%20sin27)
![\tau = 7.63 Nm](https://tex.z-dn.net/?f=%5Ctau%20%3D%207.63%20Nm)
Answer:
Explanation:
<em>Position is the location of the object (whether it's a person, a ball, or a particle) at a given moment in time.</em>
<em>Displacement is the difference in the object's position from one time to another.</em>
<em>Distance is the total amount the object has traveled in a certain period of time.</em>
<em />
<em>I hope this helps!</em>
<em />
Answer:
The kinetic energy of the particle as it moves through point B is 7.9 J.
Explanation:
The kinetic energy of the particle is:
<u>Where</u>:
K: is the kinetic energy
: is the potential energy
q: is the particle's charge = 0.8 mC
ΔV: is the electric potential = 1.5 kV
Now, the kinetic energy of the particle as it moves through point B is:
![\Delta K = K_{f} - K_{i}](https://tex.z-dn.net/?f=%20%5CDelta%20K%20%3D%20K_%7Bf%7D%20-%20K_%7Bi%7D%20)
![K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J](https://tex.z-dn.net/?f=%20K_%7Bf%7D%20%3D%20%5CDelta%20K%20%2B%20K_%7Bi%7D%20%3D%201.2%20J%20%2B%206.7%20J%20%3D%207.9%20J%20)
Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.
I hope it helps you!