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3 years ago

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Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.

1 answer:

Semmy [17]3 years ago

5 0

Answer:

Explanation:

Given

time of contact between bat and ball is $t=0.00129 s$

suppose u is the incoming velocity and v is the final velocity after collision

$m=mass\ of\ ball$

Impulse exerted is given by change in momentum of the particle.

Initial momentum $P_i=m\times u$

Final momentum $P_f=m\times v$

Change in momentum $\Delta P=P_i-P_f$

Impulse $J=F_{avg}\cdot t=\Delta P$

$J=F_{avg}\cdot t=m(u-v)$

$F_{avg}=\frac{m(v-u)}{t}$

$F_{avg}=775.2\times m(v-u)\ N$

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initial velocity=u=24m/s
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$\\ \Large\sf\longmapsto v^2=576+768$

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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

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And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

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v is the velocity of the charge

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So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

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Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

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$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

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antoniya [11.8K]

Answer:

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