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3 years ago

7

A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t

he distance from the probe to Earth?

1 answer:

adoni [48]3 years ago

5 0

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

Distance = 3.69 × 10^9 m

Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

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The weight of the body varies from place to place but the inertia of the body remains the same. Why?

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<u>Explanation: </u>

The weight as already known well, is not a universal constant, whereas the mass is. As the weight is a product of mass of the body and the gravitational acceleration working on it, it varies as the factor of acceleration involved makes it a vector quantity.

And thus be dependent upon the direction too. Inertia of the body is an independent factor, on the other hand, an inherent property and so does not change.

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3 years ago

A person is riding a bicycle, and its wheels have an angular velocity of 13.1 rad/s. Then, the brakes are applied and the bike i

lana [24]

First, let's convert the displacement to radians:

$10.9\text{ rev}=10.9\cdot2\pi\text{ rad}=21.8\pi\text{ rad}$

Now, let's calculate the angular acceleration using Torricelli's equation:

$\begin{gathered} V^2=V_o^2+2\cdot a\cdot d\\ \\ 0=13.1^2+2\cdot a\cdot21.8\pi\\ \\ 43.6\pi a=-171.61\\ \\ a=\frac{-171.61}{43.6\pi}\\ \\ a=-1.253\text{ rad/s^^b2} \end{gathered}$

Now, to calculate the time, we can use the formula below:

$\begin{gathered} V=V_o+at\\ \\ 0=13.1-1.253t\\ \\ 1.253t=13.1\\ \\ t=\frac{13.1}{1.253}\\ \\ t=10.45\text{ s} \end{gathered}$

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1 year ago

What is the method of charging used when you run a straw with fur

Orlov [11]

Answer:

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Explanation:

Friction is another method of making a charge become higher than the other.

With either straw or a fur,charging by friction is used.

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3 years ago

Which of the following is not true?

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3 years ago

Read 2 more answers

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago