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torisob [31]
3 years ago
7

Which "sphere" does an astronomer most directly study?

Physics
2 answers:
Allushta [10]3 years ago
7 0

Answer:

exosphere

Explanation:

exosphere is space

astronomy is space

Lisa [10]3 years ago
5 0
I would choose B. exosphere. Astronomers study outer space, planets and ect.
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The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a
umka21 [38]
Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
v_{max} = \omega A
Therefore
v(t) = -V_{max} sin(\omega t)

The KE (kinetic energy) is given by
KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)

The PE (potential energy) is given by
PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)

When the KE and PE are equal, then
v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)

For the oscillating spring,
\omega ^{2} =  \frac{k}{m} \\ V_{max} = \omega A =  \sqrt{ \frac{k}{m} } A
Therefore
v^{2} =  \frac{k}{m}  \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max}  \,cos( \sqrt{ \frac{k}{m} t} )

Answer: v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )

3 0
4 years ago
Sorry relized it wasnt a good picxD 4 please WILL MARK BRANLIEST
Andrej [43]
I think Y is the answer to number 4
7 0
4 years ago
A spider is crawling on a wall. First it crawls 1 meter up, then 1 meter to the left, and then 1 meter down. what is its total d
maria [59]
1 meter up, 1 meter to the left and then 1 metter down.
It goes 1 meter up and 1 meter down, so its in the same place. The only displacement is 1 meter to the left.

Look at picture:

(hope this helps)

4 0
3 years ago
Read 2 more answers
Man has never directly reached the inner layers of the Earth. But what phenomena allow to determine its chemical composition?
AnnyKZ [126]
I think letter A.Surface wave propagation
7 0
3 years ago
A ball is thrown 24 m/s into the air. How high does it go?
Ierofanga [76]

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

5 0
3 years ago
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