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GREYUIT [131]
3 years ago
9

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite

direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Physics
1 answer:
nalin [4]3 years ago
7 0

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s

Hence, this is the required solution.

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A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

4 0
3 years ago
A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori
Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

       Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.

That's the force that's accelerating the little block, so that must be the tension
in the string.


7 0
3 years ago
How do galaxies vary?
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4 0
3 years ago
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6 0
3 years ago
Read 2 more answers
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