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Luden [163]
4 years ago
6

One light-year is the distance light travels in one year. This distance is equal to 9.461 1015 m. After the sun, the star neares

t to Earth is Alpha Centauri, which is about 4.35 light-years from Earth. Express this distance in a. megameters. b. picometers.
Physics
1 answer:
Oliga [24]4 years ago
4 0

Answer:

A.) 411.6 Mega metres

B.) 411.6 × 10^18 Picometers

Explanation:

Given that One light-year is the distance light travels in one year. This distance is equal to 9461 1015 m.

After the sun, the star nearest to Earth is Alpha Centauri, which is about 4.35 light-years from Earth.

The distance travelled will be

Distance = 9461 1015 × 4.35

Distance = 411557915.3 m

To Express this distance in

a. megameters, divide the value by 1000000

411557915.3 / 1000000

411.5579153 Mega metre

411.6 Mega metres approximately

b. picometers, multiply by 10^12

That is, 411557915.3 × 10^12

411.6 × 10^18 Picometers

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bazaltina [42]

The centripetal force : F = 293.3125 N

<h3>Further explanation</h3>

Given

mass = 65 kg

v = 9.5 m/s

r = 20 m

Required

the centripetal force

Solution

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle  

\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}

F = centripetal force, N  

m = mass, Kg  

v = linear velocity, m / s  

r = radius, m  

Input the value :

F = 65 x 9.5² / 20

F = 293.3125 N

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a toy car was on the floor in the play room until the child applied force by starting to race with it. it accelerated and proceeded to move at 5m/s east. it came to a stop when the bedpost applied force to it.

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Which of the following is not used to measure wind?
lapo4ka [179]

Answer:

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Explanation:

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4 years ago
Read 2 more answers
A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the
Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

r = \sqrt{(3.5)^2+(2.5)^2}

  =\sqrt{8.5}

SL=85

As we know,

⇒  SL=10 \ log(\frac{I}{I_o} )

     85=10 \ log(\frac{I}{10^{-12}} )

      I=3.162\times 1^{-4} \ W/m^2

Now,

⇒  P_{front} = I(2\pi r^2)

               =(3.162\times 10^{-4})(2\pi\times 18.5)

               =0.0368 \ W

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                =0.049 \ W

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22= 1/2 9.8 t^2
djverab [1.8K]

Answer:

5790 i dony kiow

Explanation:

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