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Eva8 [605]
3 years ago
12

In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it l

eaves his bat with a speed of 20.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball
Physics
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

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Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se
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Answer:

R = 9.85 ohm , r = 0.85 ohm

Explanation:

Let the two resistances by r and R.

when they are connected in series:

V = 12 V

i = 1.12 A

The equivalent resistance when they are connected in series is

Rs = r + R

So, By using Ohm's law

V = i Rs

Rs = V / i = 12 / 1.12 = 10.7 ohm

R + r = 10.7 ohm    .... (1)

When they are connected in parallel:

V = 12 V

i = 9.39 A

The equivalent resistance when they are connected in parallel

R_{p}=\frac{R+r}{rR}

So, By using Ohm's law

V = i Rp

Rp = V / i = 12 / 9.39 = 1.28 ohm

\frac{R+r}{rR}=1.28    .... (2)

by substituting the value of R + r from equation (1) in equation (2), we get

r R = 8.36 ..... (3)

R-r = \sqrt{\left ( R+r \right )^{2}-4rR}

R-r = \sqrt{\left ( 10.7 \right )^{2}-4\times 8.36}=9 ..... (4)

By solvng equation (1) and (4), we get

R = 9.85 ohm , r = 0.85 ohm

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