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11111nata11111 [884]
3 years ago
6

What is the circumference of the moon?

Physics
2 answers:
Fittoniya [83]3 years ago
8 0

The diameter of the moon is about 2,161 miles or (3,476 kilometers).that makes the circumference about 6,790 miles

iren [92.7K]3 years ago
7 0
The circumference of the moon is 6,786 mi
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If the wavelength of a wave increases does its frequency also increase?
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Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accele
sveta [45]

Answer:

The time that passes until the police catch the speeder is 82.6204 seconds.

Explanation:

A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.

The position is calculated by the expression:

x = x0 + v0*t + 1/2*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

First, let’s look at the police car’s equations of motion. In this case:

  • x0= 0
  • v0= 50 m/s
  • a= 2 m/s²

So: x = 50 m/s*t + 1/2*2 m/s²*t²

Now for the speeder’s car’s equations of motion you know:

  • x0= 3 km= 3,000 m
  • v0= 55 m/s
  • a= 1 m/s²

So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

When the police catch the speeder they are both in the same position. So:

50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

Solving:

0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²

0= 3,000  + 55 *t + 1/2*t² - 50*t - 1*t²

0= 3,000  + 55 *t - 50*t - 1*t² + 1/2*t²

0= 3,000  + 5*t - 1/2*t²

Applying the quadratic formula:

x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000  } }{2*(-\frac{1}{2} )}

x1= -72.6209

and x2= 82.6209

Since you are calculating the value of a time and it cannot be negative, then <u><em>the time that passes until the police catch the speeder is 82.6204 seconds.</em></u>

8 0
3 years ago
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