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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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Visual map interpretation is usually aided by the use of legends. The legend hold the key to the elements which are represented on the map. Usually, a combination of colors and shapes makes up the legend and makes map interpretation easy.

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Learn more : brainly.com/question/25323763

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2 years ago
I can use everything on my body to advance the ball but what?
ddd [48]
The answer would be hands! hope this helps
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2 years ago
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Lelu [443]
What do you need help on
4 0
3 years ago
A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of
marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0
3 years ago
Which of the following are required for scientific questions
icang [17]
The answer is D-Testable

Hope this helps
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3 years ago
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