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oee [108]
2 years ago
12

If Bill threw a ball straight up on the Moon (g=1.6 m/s2) with a starting velocity of 22m/s from a cliff and it fell past him an

d reached a velocity of 56m/s as it impacted the Moon’s surface below the cliff, how long was the flight time?
Physics
1 answer:
kap26 [50]2 years ago
5 0

Answer:

Write two important of physical state

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f biofuel production and use reduces our consumption of imported fossil fuels, we may become less vulnerable to the adverse impacts of supply

Explanation:

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3.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s . Its total moment of inertia is 1710 kg
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When we look at Jupiter, we're actually seeing the outermost layer of its clouds.

The Great Red Spot is a storm in Jupiter's southern hemisphere with crimson-colored clouds that spin counterclockwise at wind speeds

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Uranus' atmosphere is mostly hydrogen and helium, with a small amount of methane and traces of water and ammonia.

As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.

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Neptune has an average temperature of -353 Fahrenheit (-214 Celsius).

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5 0
3 years ago
A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief
fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

x_1 = -120cos70^o = -41.04 km

y_1 = -120sin70^o = -112.763 km

After the 2nd flight its coordinate would be:

x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km

y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km

So in order to fly back to its HQ it must fly a distance and direction of

s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km

tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71

\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o north of east

3 0
3 years ago
A satellite with a mass of 110 kg and a kinetic energy of 3.0 ´ 109 J must be moving at a speed of
djverab [1.8K]

Answer: A satellite with a mass of 110 kg and a kinetic energy of 3.08×10^9 J must be moving at a speed of 7483 m/s.

Explanation: To find the answer we need to know about the kinetic energy of a body.

<h3>How to solve the problem the equation of kinetic energy?</h3>
  • We have the expression for kinetic energy of a body as,

                                   KE=\frac{1}{2}mv^2

  • Given that,

                                   m=110kg\\KE=3.08*10^9J\\

  • We have to find the speed of the satellite,

                               v=\sqrt{\frac{2KE}{m} } =\sqrt{\frac{2*3.08*10^9}{110} } =7.483*10^3 m/s

Thus, we can conclude that, the velocity of the satellite will be 7438m/s.

Learn more about Kinetic energy here:

brainly.com/question/28105739

#SPJ4

8 0
2 years ago
Read 2 more answers
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