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3 years ago

10

How far away is the furthest artificial satellite orbiting earth?

1 answer:

rusak2 [61]3 years ago

8 0

Real-time distance and velocity data is provided by NASA and JPL. At a distance of 153.88 AU (23.020 billion km; 14.304 billion mi) from Earth as of September 5, 2021, it is the most distant artificial object from Earth. The probe made successful flybys of Jupiter, Saturn, and Saturn's largest moon, Titan.

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The part of the computer that provides access to the internet is

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The Modem

or physically connecting on the computer the data port is for a cable to connect

5 0

3 years ago

Please help on this one

Yuki888 [10]

The answer is c because I had the same thing and it was right

5 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

Indicate whether the statement is true or false. an astronaut weighs the same on earth as in space.

tino4ka555 [31]

Weight is different (but mass is the same)

6 0

3 years ago

Read 2 more answers

Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= $\sqrt{(8)^2+(-2)^2}$

|v|= $\sqrt{68}$ = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0

3 years ago