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Reptile [31]
3 years ago
9

A ball traveling with an initial momentum of 4.5 kgm/s bounces off a wall and comes back in the opposite direction with a moment

um of –3.5 kgm/s. What is the change in momentum?
Physics
1 answer:
irinina [24]3 years ago
3 0

Answer:

-7.5 kg m/s

Explanation:

Initial momentum (p1) = 4.5 kg m/s

final momentum (p2) = -3.5 kg m/s

We have to find out the change in momentum , As linear momentum is a vector quantity. So, we have to use the vector form to solve the momentum change . Means we have to take the sign of momentum also.

Change in momentum = Final momentum - initial momentum

So,

Change in momentum = p2 - p1

Insert values from question,

Change in momentum = -3.5 kg m/s - 4 kg m/s

So, change in momentum = -7.5 kg m/s

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BFSkinner enfatizó la importancia de   creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.

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Which scenario describes a systematic error?
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A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical c
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<h2>This vertical component of velocity is 35 m/s </h2>

Explanation:

The dart is project with 150 m/s from a point at an angle of 30⁰

The vertical component of velocity = 150 sin 30 = 75 m/s

Thus initial vertical velocity is = 75 m/s

The velocity after 4 s can be calculate by

v = u - g t

here u is the initial velocity and t is the time , g is the acceleration due to gravity .

Thus v = 75 - 10 x 4 = 35 m/s

6 0
3 years ago
n an oscillating LC circuit, L = 3.76 mH and C = 3.13 μF. At t = 0 the charge on the capacitor is zero and the current is 2.95 A
abruzzese [7]

Answer:

Part a)

Q = 320 \mu C

Part b)

t = 8.52 \times 10^{-5} s

Part c)

Rate of energy = 301.5 J/s

Explanation:

Part a)

Since energy is always conserved in LC oscillating system

So here for maximum charge stored in the capacitor is equal to the magnetic field energy stored in inductor

\frac{1}{2}Li^2 = \frac{Q^2}{2C}

now we have

Q = \sqrt{LC} i

Q = \sqrt{(3.76 \times 10^{-3})(3.13 \times 10^{-6})} (2.95)

Q = 320 \mu C

Part b)

Energy stored in the capacitor is given as

U = \frac{q^2}{2C}

now rate of energy stored is given as

\frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt}

so here we also know that

q = Q sin(\omega t)

\frac{dq}{dt} = Q\omega cos(\omega t)

now from above equation

\frac{dU}{dt} = \frac{Qsin(\omega t)}{C} (Q\omega cos\omega t)

so maximum rate of energy will be given when

sin\omega t = cos\omega t

\omega t = \frac{\pi}{4}

t = \frac{\pi}{4}\sqrt{LC}

t = 8.52 \times 10^{-5} s

Part c)

Greatest rate of energy is given as

\frac{dU}{dt} = \frac{Q^2\omega}{C}

\frac{dU}{dt} = \frac{(320 \mu C)^2 \sqrt{\frac{1}{(3.76 mH)(3.13 \mu C)}}}{3.13 \mu C}

\frac{dU}{dt} = 301.5 J/s

7 0
4 years ago
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