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3 years ago

8

Express the distance between the moon and earth in meters with a meter prefix (km).

2 answers:

valkas [14]3 years ago

5 0

Answer:

$0.384 Gm$

Explanation:

The average distance between the moon and earth is 384,400 km

Why average distance? Well, the Moon is not always the same distance from Earth, its orbit is not a perfect circle.

Expressing this distance in meters we have:

$384000km*\frac{1000m}{1km}=0.384*10^9m$

Since 1 gigameter (Gm) is $10^9 m$ :

$0.384*10^9m=0.384 Gm$

tankabanditka [31]3 years ago

3 0

Answer;

= 3.86 × 10^8 Meters

Explanation;

-The distance between the Earth and the moon is 386000 km

But; 1 km = 1000 m

Therefore; 386000 km will be equivalent to;

= 386000 × 1000

= 386000000 m

= 3.86 × 10^8 meters

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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

=> $S = 24.86 ft$

5 0

3 years ago

The closeness of measured values to an accepted value is the BLANK of the data ?

wlad13 [49]

When you refer to how close a measured value is to a standard, accepted or known value, you are talking about the ACCURACY of the data. This is the definition of accuracy when it comes to engineering and other fields of science.

Accuracy is usually associated or with the term precision, as their definitions are often interchanged.

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3 years ago

A) explain how the data matches each line on the graph (give name and line letter for each)

stepladder [879]

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2 years ago

The way light bounces off a minerals surface is described by the minerals?

Step2247 [10]

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3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago