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timofeeve [1]
3 years ago
8

Radio signals travel at a rate of 3 × 108 meters per second. how many seconds would it take for a radio signal to travel from a

satellite to the surface of earth if the satellite is orbiting at a height of 9.6 × 106 meters? (hint: time is distance divided by speed.)
Physics
1 answer:
nlexa [21]3 years ago
6 0
<span>3.2x10^-2 seconds (0.032 seconds)
   This is a simple matter of division. I also suspect it's an exercise in scientific notation, so here is how you divide in scientific notation:

   9.6 x 10^6 m / 3x10^8 m/s

   First, divide the significands like you would normally.
 9.6 / 3 = 3.2

   And subtract the exponent. So
 6 - 8 = -2

   So the answer is 3.2 x 10^-2
 And since the significand is less than 10 and at least 1, we don't need to normalize it.

   So it takes 3.2x10^-2 seconds for the radio signal to reach the satellite.</span>
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Answer:

29 m/s.

Explanation:

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What force is needed to propel a 421-kg car with an acceleration of 3.77 m/s2?
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Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

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    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

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    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

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3 years ago
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Answer:

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