Physics, 29.06.2019 22:00 joeljuarez128oyusun

A95 kg fullback is running at 3.0 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west. calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.

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Answer from: mya9859

(a) The original momentum of the fullback is 285 kgm/s.

(b) The impulse exerted on the full back is -285 kgm/s.

(d) The average force exerted on the tackler is 356.25 N.

The given parameters;

mass of the fullback, m = 95 kginitial velocity of the fullback, u = 3 m/stime of motion of the fullback, t = 0.85 s

The original momentum of the fullback is calculated as follows;

$P_i = mv\\\\P_i = 95 \times 3\\\\P_i = 285 \ kg.m/s$

The impulse exerted on the full back is calculated as follows;

$J = \Delta P = mv_f - mv_i\\\\J = m(v_f - v_i)\\\\J = 95(0-3)\\\\J = - 285 \ kg.m/s$

The impulse exerted on the tackler is calculated as follows;

$J_1 = -J_2\\\\J_2 = -J_1\\\\J_2 = -(-285 \ kgm/s)\\\\J_2 = 285 \ kgm/s$

The average force exerted on the tackler is calculated as follows;

$F = \frac{\Delta mv}{t} \\\\F = \frac{285}{0.8} \\\\F = 356.25 \ N$

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Answer from: leilakainani26

(a) Original momentum of the fullback: 285 kg m/s

The original momentum of the fullback is equal to the product between his velocity and its mass:

p_i=mv

where:

m = 95 kg is the mass

v = 3.0 m/s is the velocity

Substituting the numbers into the formula, we find

p_i=(95 kg)(3.0 m/s)=285 kg m/s

(b) Impulse exerted on the fullback: -285 kg m/s

The impulse exerted on the fullback is equal to his variation of momentum:

$I=\Delta p=p_f -p_i$

where:

p_f = 0 is the final momentum of the fullback (zero because he comes to a stop)

p_i = 285 kg m/s is the initial momentum

Substituting,

I=0-(285 kg m/s)=-285 kg m/s

The total momentum of the fullback and the tackler must be conserved:

p_i + P_i = p_f + P_f

where p_i and p_f are the initial and final momentum of the fullback, while P_i, P_f are the initial and final momentum of the tackler.

We can re-arrange the equation as follows:

$P_f - P_i = p_i -p_f\\\Delta p_{tackler} = -\Delta p_{fullback}\\I_{tackler} = -I_{fullback}$

which means that the impulse exerted on the tackler is the negative of the impulse exerted on the fullback, so:

I=-(-285 kg m/s)=285 kg m/s

(d) Average force exerted on the tackler: 335.3 N

The impulse on the tackler is equal to the product between the average force and the time of the collision:

$I=F \Delta t$

Since $\Delta t=0.85 s$ , we can find the average force:

$F=\frac{I}{\Delta t}=\frac{285 kg m/s}{0.85 s}=335.3 N$

Answer from: Quest

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )

Answer from: Quest

Maybe c? sorry if im wrong
An object is located 5.0 cm from a convex lens. the lens focuses light at a distance of 10.0 cm. wha

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