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horrorfan [7]
4 years ago
9

What does the churning air in the troposphere help determine

Physics
1 answer:
bogdanovich [222]4 years ago
8 0
The churning air in the troposphere helps determine the altitude of a place
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The crate is at rest on the incline. What can you say about the force of friction acting on the crate?A. the frictional force po
netineya [11]

Answer:

The frictional force points up the incline.

Explanation:

It is given that, the crate is at rest on the incline.  We know that the forces acting on the crate are force of gravity, normal and the force of friction. We know that the frictional force is the force that opposes the motion of an object.

In this case the crate is at rest. The crate will have static friction, that points up the incline.

So, the correct option is (A) "the frictional force points up the incline".

7 0
3 years ago
IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U
Bingel [31]
<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

Cold air is dense and sinks to the surface to create what is
known as an area of high pressure.</span>


8 0
4 years ago
Read 2 more answers
A 76 N crate is hung from a spring
lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0
3 years ago
Since the moon orbits the earth approximately every 27 days, there is a lunar eclipse approximately every 27 days.
Dahasolnce [82]

Answer:

No ! its False

Explanation:

We all know , we dont have lunar eclipse in every 27 days . They do not happen every month because the Earth's orbit around the sun is not in the same plane as the Moon's orbit around the Earth.

5 0
3 years ago
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made
anyanavicka [17]

Answer:

Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

Explanation:

Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2}  } }      ......(1)

Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

In this problem,

Current, I = 0.7 A

Length of wire, L = 0.62 m

Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

Substitute these values in equation (1).

B=\frac{4\pi\times10^{-7}\times  0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2}  } }

B = 6.99 x 10⁻⁶ T

3 0
3 years ago
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