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2 years ago

8

The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo

tball fields (length=91.4 m) does the shuttle cover in the blink of an eye? The three-toed sloth is the slowest-moving land mammal. On the ground, the sloth moves at an average speed of 0.033 m/s, considerably slower than the giant tortoise have gone relative to the sloth? Do not put units in your answer.

1 answer:

sveta [45]2 years ago

7 0

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

$Distance=Speed\times Time$

Thus distance covered in blinking of eye =

$Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters$

Thus no of football fields= $\frac{936}{91.4}=9.15Fields$

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2 years ago

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2 years ago

What is the equivalent resistance of the

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Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

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Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .

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1 year ago

Explain what is meant by “In Phase” and “Out of Phase?

LekaFEV [45]

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2 years ago

Read 2 more answers

Two speakers are spaced 15 m apart and are both producing an identical sound wave. You are standing at a spot as pictured. What

IgorLugansk [536]

Answer:

Frequency is $213.04\ s^{-1}.$

Explanation:

Distance between source 1 from the receiver , $S_1 =\sqrt{10^2+22^2}=24.17\ m.$

Distance between source 2 from the receiver , $S_2=\sqrt{5^2+22^2}=22.56\ m.$

Now ,

Path difference , $r = S_1-S_2=24.17-22.56=1.61\ m.$

We know, for constructive interference path difference should be integral multiple of wavelength .

Therefore, $r=n\times \lambda$

It is given that n = 1,

Therefore, $\lambda=1.61\ m.$

Frequency can be found by , $\nu=\dfrac{v}{\lambda}= \dfrac{343}{1.61}= 213.04\ s^{-1} .$

Hence, this is the required solution.

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2 years ago