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Korvikt [17]
3 years ago
7

Kepler's____ of planetary motion states that the planets have an elliptical orbit, with the Sun at one focal point of the ellips

e. A. 1st law B. 2nd law C. 3rd law
Physics
1 answer:
pentagon [3]3 years ago
6 0
A.Kepler's first law of planetry motion.
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After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187
Dmitry_Shevchenko [17]

The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

v=\frac{d}{t}

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial  time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

t = 2.5 h \cdot (60)(60)=9000 s

Therefore, the average speed for the trip is

v=\frac{187,000}{9000}=20.8 m/s

Learn more about average speed:

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3 years ago
An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega
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An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>
  • A particle that has a negative angular velocity rotates counterclockwise.
  • Negative angular acceleration () is a "push" that is hence counterclockwise.
  • The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
  • For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
  • If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
  • It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

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In 10s a car accelerates 4m/s^2 to 60m/s. How fast was the car going before it accelerated?
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v2 = v1 + at 60 = v1 + 4*10  v1 = 20 m/s

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