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Hoochie [10]
3 years ago
7

A car moving in a straight line starts at x=0 at t=0. it passes the point x=25.0m with a speed of 11.0m/s at t=3.0s. it passesth

e point x=385m with a speed of 45.0m/s at t=20.0s. find the average velocity and the average acceleration, between t=3.0s and t=20.0s
Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Between time 3 second to 20 seconds the time difference = 17 seconds

Distance traveled = 385 - 25 = 360 m

So average velocity = 360/17 = 21.18 m/s

Change in velocity at t =3 seconds and t = 20 seconds is given by 45 - 11 = 34 m/s

So average acceleration = 34/17 = 2  m/s

         

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A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distan
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Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i  - I \omega_f \\\\m = \frac{ I \omega_i  - I \omega_f}{r^2 \omega_f }

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3 years ago
Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
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To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

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h = Height

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A_1 = 2.4*10^{-4} m^2

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h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

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A_1v_1 = A_2v_2

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A_2= \frac{A_1v_1 }{v_2}

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A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

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