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2 years ago

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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

hat will his resulting velocity be?

1 answer:

AysviL [449]2 years ago

3 0

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

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A beam of unpolarized light shines on a stack of five ideal polarizers, set up so that the angles between the polarization axes

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Answer:

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Explanation:

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$\theta$ = Angle between each pair of adjacent polarizers

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It is given that

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we know that the intensity of light after passing through "n" polarizers is given as

$I_{n} = (0.5) I_{o} Cos^{2n-2}\theta$

$\frac{I_{n}}{I_{o}} = (0.5) Cos^{2n-2}\theta$

inserting the values

$0.277 = (0.5)Cos^{2(5)-2}\theta$

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$Cos\theta = 0.9288$

$\theta = Cos^{-1}(0.9288)$

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Explanation:

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Net force on the lower box

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Considering force on the upper box

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Put this value of a in the equation above

T - m gμ_s = M g μ_s

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2 years ago

Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

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The average speed is defined by the variation of the position between the time spent

v = Δx / Δt

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