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zheka24 [161]
3 years ago
13

How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?

Physics
1 answer:
Sonja [21]3 years ago
6 0

Answer:

Explanation:  We have seasons because Earth's axis – the imaginary line that goes through the Earth and around which the Earth spins — is tilted. It's tilted about 23.5 degrees relative to our plane of orbit (the ecliptic) around the Sun. As we orbit our Sun, our axis always points to the same fixed location in space. Our northern axis points almost directly toward Polaris, the North Star.

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if a man has a mass of 83 kilograms on earth, what will the force of gravity on his body be on the moon
AnnZ [28]

If a man has a mass of 83 kilograms on Earth, the force of gravity on his body be on the moon 135.6N. force =mass*acc , 83 * 9.8/6= 813.4/6 = 135.6N

3 0
3 years ago
Read 2 more answers
How much force is needed to lift a 25-kg mass at a constant verlocity?
Troyanec [42]

-- In order to achieve constant verlocity, the net force on the mass must be zero.  So if there ARE any forces acting on it, they must be balanced.

-- There is already a force on the mass that can't be eliminated . . . the force of gravity.

-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = <em><u>245N</u></em> in the <u><em>downward</em></u> direction.

-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of <em>245N</em>, all in the <em>upward</em> direction.

-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.

5 0
3 years ago
Read 2 more answers
The spool has a mass of 50 kg and a radius of gyration of ko = 0.280 m. if the 20-kg block a is released from rest, determine th
Rzqust [24]
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis 
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is 
T = 1/2 I₀w² + 1/2mAVA²

= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation  is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
5 0
3 years ago
Provide descriptions and examples of the following forces: friction, gravity and elastic.
arsen [322]

Answer:

Friction:-

The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing inter molecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an

EG:-

A coaster sliding against a table.

Gravity:-

The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found

EG:-

The force that causes a car to coast downhill even when you aren't stepping on the gas.

Elastic:-

Elasticity is the ability of a material to return to its original shape after being stretched or compressed. When an elastic material is stretched or compressed, it exerts elastic force. This force increases the more the material is stretched or compressed.

EG:-

An archer's stretched bow

5 0
3 years ago
Suppose that placing 0.3 inch of lead in front of a gamma source reduces the count rate from 1045 cps to 573 cps. What is um^-1
Ainat [17]

Answer:

14.49 g/cm²

Explanation:

I = Io e^-(ux)

Where:

I = 573

Io = 1045

x = 0.3 inches and

rho = 11.4g/cm^3

Using the conversion constant

1 inch = 2.54 cm;

0.3 inches = 0.3 * 2.54 cm

0.3 inches = 0.762 cm

I/Io = e^-(ux), or say

Io/I = e^(ux), taking the In of both sides

ln(Io/I) = ux, making u subject of formula

u = 1/x * ln(Io/I)

u = 1/0.762 * ln(1045/573)

u = 1.312 * 0.6

u = 0.787

Next, we say that

u/rho = 0.7872/11.4 = 0.069

And finally, we make

1/(u/rho) to be our final answer

Inverse of the answer is = 14.49 g/cm²

Therefore, the um^-1 in g/cm^2? is 14.49

5 0
2 years ago
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