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mel-nik [20]
2 years ago
5

The __________ is the layer of the atmosphere where satellites and the space shuttle orbit the Earth.

Physics
2 answers:
Y_Kistochka [10]2 years ago
7 0

Answer: d. exosphere

The exosphere is the outermost layer of the atmosphere of earth. It is present above the thermosphere which is lying above the earth. From this layer, the atoms and molecules escape out, which maintains an environment for the movement of satellites and space shuttles.


Gemiola [76]2 years ago
3 0
The most likely answer, according to an informational article I have read, on the earths atmosphere, is exosphere. Hope it helps! If u need more info, message me!
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A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
2 years ago
What is the new kinetic energy of the 1900 kg ship on the right moving at 4 m/s?
Elis [28]

Answer:

soi nuevo

Explanation:me  regalan puntos

7 0
3 years ago
Read 2 more answers
If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step
azamat

Answer:

Explanation:

v_f^2 = v_i^2-2a(x_f-x_i)

Subtract both sides by v_i^2:

- v_i^2+v_f^2 = -2a(x_f-x_i)

Divide both sides by -2*a:

\frac{v_i^2 - v_f^2}{2a} =x_f-x_i

Add both sides by x_i:

x_i+\frac{v_i^2 - v_f^2}{2a} =x_f

Subtract both sides by \frac{v_i^2 - v_f^2}{2a}:

x_i=x_f-\frac{v_i^2 - v_f^2}{2a}

8 0
3 years ago
A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
aivan3 [116]

Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

F =20.0N

Theta = 37°

W = FSCos(theta)

W = 20×6.0×Cos(37)

W = 95.8J

8 0
2 years ago
Read 2 more answers
Work done depends on
natima [27]

Answer:

C. Both force and displacement

Explanation:

Hope this helps

3 0
2 years ago
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