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Dennis_Churaev [7]
3 years ago
14

Please help !!!! I’m taking a quiz

Physics
1 answer:
icang [17]3 years ago
7 0

Answer: 512 and 16

Explanation:

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a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance
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The car's (average) acceleration would be

a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}

The car's position over time would be given by

x=v_0t+\dfrac12at^2

so that after 2.4 seconds, the car will have traveled a distance of

x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2

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3 years ago
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4 years ago
A ball is thro.d hit the grouvelocity)?
kati45 [8]

Answer:

The velocity of the ball is 3.52 m/s.

Explanation:

A projectile is any object that moves under the influence of gravity and momentum only. Examples are; a thrown ball, a fired bullet, a kicked ball, thrown javelin, etc.

Given that the ball was thrown vertically upward on the top of a skyscraper of  height 61.9 m. So that the velocity can be determined by;

      u = \sqrt{\frac{2H}{g} }

Where: u is the velocity of the object, H is the height and g is the gravitational force on the object. Given that: H = 61.9 m and g = 10 m/s^{2}, then;

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u  = 3.5185

The velocity of the ball is 3.52 m/s.

3 0
4 years ago
A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 270 kg · m2 and is rotating at 8.0 rev/min about a
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Answer:

\dot n = 6.042\,rpm

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The final angle speed of the merry-go-round is determined with the help of the Principle of Angular Momentum Conservation:

(270\,kg\cdot m^{2})\cdot \left(8\,rpm\right) = [270\,kg\cdot m^{2}+(27\,kg)\cdot (1.80\,m)^{2}]\cdot \dot n

\dot n = 6.042\,rpm

3 0
3 years ago
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