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Maru [420]
3 years ago
13

Once juno reaches jupiter, what is the minimum amount of time it takes for the transmitted signals to travel from the spacecraft

to earth?
Physics
2 answers:
arsen [322]3 years ago
5 0

speed of light = juno to earth/time

time= juno/3x10^8. if dist in metres answer is in seconds.

probably not long

Komok [63]3 years ago
4 0

In very very very round figures . . .

-- Jupiter is about 5.2 times as far from the sun as the earth is.

-- So when Jupiter and the EARTH are aligned in both orbits, Jupiter is about

(4.2) x (150 million kilometers) = 630 million kilometers

Time = (distance) / (speed)

The speed of light and radio is 300,000 km/second

Time = (630 million / 300 thousand)

<em>Time = 2,100 seconds</em>

That's 35 minutes.

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Which of these words describes humans
MA_775_DIABLO [31]
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3 0
3 years ago
A sprinter runs for 6.45 s at 7.44 m/s. How far does she get?
Gwar [14]

Answer:

D is the answer

Explanation:

6.45×7.44= 47.98800

Which if we round of we get 48m

8 0
3 years ago
It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0
4 years ago
What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

So

P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa

So The Force is given as:

F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N

The Force act on each side is 2.43×10⁴N

3 0
3 years ago
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