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stiv31 [10]
3 years ago
13

Friction and air resistance are forces that always:

Physics
1 answer:
Softa [21]3 years ago
7 0
2 act in the opposite direction. To an objects motion, tending to slow it down
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PLEASE HELP 100 POINTS!!!!!
Bas_tet [7]

Answer:

C

Explanation:

Carbon dioxide should be the input, and oxygen should be the output.

3 0
3 years ago
Read 2 more answers
Would you be doing any more work by going up the stairs twice as fast?
Ivanshal [37]
No, you wouldn't do any more work by going up the stairs twice as fast.

work done is equal to : F x D

time is not included, which mean you only done the work faster, but not doing any more work compared to the slower one

hope this helps
4 0
3 years ago
PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!
In-s [12.5K]

Answer:

I'm no expert, but multitasking is a no bueno. Because multitasking reduces your efficiency and performance because your brain can only focus on one thing at a time. When you try to do two things at once, your brain lacks the capacity to perform both tasks successfully.

Therefore I would choose A, considering he could easily workout another time, whilst the test is a one time thing. He can't redo the test, but he can workout again. So my answer is A.

If he HAS to workout and study at the same time, then C would be appropriate.

I hope this helps!

5 0
3 years ago
A person shouting at the top of his lungs emits aboue 1.0 w of energy as sound waves. What is the sound intensity 1.0 m from suc
umka21 [38]

Answer:

I=0.0795\ W/m^2

Explanation:

When a person shouts it emits 1 W of energy as sound waves, P = 1 W

Let I is the sound intensity 1 meters from such a person. We know that the power per unit area is called the sound intensity of the person. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{1\ W}{4\pi (1\ m)^2}  

I=0.0795\ W/m^2

So, the sound intensity of such a person is 0.0795\ W/m^2. Hence, this is the required solution.

3 0
4 years ago
In the future, an experimental spacecraft leaves space dock for a test flight. After flying in a very large circle at constant s
Fantom [35]

Answer:

2.58 x 10⁸ m/s

Explanation:

Time dilation fomula will be applicable here, which is given below.

t = \frac{T}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}

Where T is dilated time or time observed by clock in motion , t is stationary time , v is velocity of clock in motion and c is velocity of light .

c is 3 times 10⁸ ms⁻¹ , T is 7.24 h , t is 3.69 h. Put these values in the formula

7.24 = \frac{3.69}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}\\

\frac{v^2}{c^2}=0.744\\\\

v=2.58\times 10^8

5 0
3 years ago
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