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timofeeve [1]
3 years ago
12

The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

of the orbit and shaded segments are the same?
Physics
1 answer:
strojnjashka [21]3 years ago
4 0

Answer: not sure

Explanation:

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While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from
Alona [7]

Answer: FM radio waves, AM radio waves, yellow light, micro waves

Explanation:

5 0
3 years ago
Read 2 more answers
If a bullet travels at 593.0 m/s, what is its speed in miles per hour?
Ksenya-84 [330]

We have 1 \; mile = 1609.34\; meters. So, 1 \; meter = \frac{1}{1609.34} \;mile.

1 \; hour = 3600 \; s. So 1 \; s = \frac{1}{3600}  \; hour.

Thus we can convert the units of the given quantity.

That is,

593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour.

The quantity is converted to the required units.


7 0
2 years ago
How can you determine the amount of work done on an object? (joules, work, power...etc.)
Novay_Z [31]
Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
7 0
2 years ago
A spaceprobe has an 29.0 m length when measured at rest. What length
elixir [45]

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let L_0 be the length of the space-probe when measured at rest, and L be its length as observed by an observer moving at velocity v, then

(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }

Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

3 0
3 years ago
The third one pls help
Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 0
3 years ago
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