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Sholpan [36]
2 years ago
8

Which layer of the atmosphere is just above the troposphere?

Physics
1 answer:
boyakko [2]2 years ago
6 0
The answer would Stratosphere because <span>1) The </span>troposphere<span> is the first </span>layer above<span> the surface and contains half of the Earth's </span>atmosphere<span>. Weather occurs in this </span>layer. 2) Many jet aircrafts fly in thestratosphere<span> because it is very stable. Also, the ozone </span>layer<span> absorbs harmful rays from the Sun</span>
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True or False: A chemical reaction always happens when two substances are combined. (please help fast this is a test)
Makovka662 [10]

Answer:

no not always sometimes they react at all so false I hope I helped :)

6 0
2 years ago
Read 2 more answers
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
Please help in this question!! I will mark brainlest​
LUCKY_DIMON [66]

Answer:

im not too sure about that all i know is history

4 0
2 years ago
What does a negative value for the focal length tell you?
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LOL SOMEONE DO THIS FOR ME!!!
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