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iren2701 [21]
2 years ago
6

A university is planning to send an optical telescope into orbit. stuents attending the university will be able to view which of

the following?
Physics
2 answers:
AysviL [449]2 years ago
6 0

Options are

A.) Microwaves coming from Earth

B.) X-rays coming from Earth

<u>C.) White light coming from space</u>

D.) Gamma rays coming from space

C would be your answer, I can't explain it but I just took the test on Plato.

kodGreya [7K]2 years ago
3 0
Where are the answers? If there's anything about a white light coming from space I would choose that one. 
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two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f
Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0
2 years ago
Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o
nevsk [136]
In the circular motion of the hammer, the centripetal force is given by
F=m \frac{v^2}{r}
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N
4 0
2 years ago
Read 2 more answers
A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?
Aleksandr [31]

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

8 0
2 years ago
Read 2 more answers
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
1 year ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
Degger [83]

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

n= \nu _{1}- \nu _{2}.

Substitute the values of frequencies of 2 tuning forks in the above equation.

n=(260 Hz-256 Hz)\\n=4

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0
1 year ago
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