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denis23 [38]
3 years ago
11

The sun radiates energy. Does the earth similarly radiate energy? If so, why can’t we see the radiant energy from the earth?

Physics
1 answer:
notsponge [240]3 years ago
6 0

Yes, the earth radiates energy, but since the peak frequency f is directly proportional to the absolute temperature of the radiator, the wavelength of the radiation is far too long for us to see.

Answer: Option C

<u>Explanation:</u>

Radiations or light rays are the basic name for electromagnetic energy packets travelling through space. It goes extremely quick (multiple times around the earth in one second) and can go through a vacuum. It needn't bother with material to travel in.

It has numerous structures, including visible light, infrared (IR), bright (UV), X-beams, microwaves, and radio waves. These are a no different structure of energy, just with various frequencies and measures of energy. Various frequencies of radiation communicate with an issue in an unexpected way, which causes them to appear to be more changed to us than they truly are.

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A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits
slavikrds [6]

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

  • lets first calculate the time it took the fish to fall

s = ut + (\frac{1}{2}) at^{2}

since u = 0

s =  (\frac{1}{2}) at^{2}

t = \sqrt{\frac{2s}{a} }

where a = acceleration due to gravity and s = vertical height

t = \sqrt{\frac{2 x 4.7 }{9.8} } = 0.98 s

  • pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

  • lets also calculate the time it will take the fish to fall

s = ut + (\frac{1}{2}) at^{2}

since u = 0

s =  (\frac{1}{2}) at^{2}

t = \sqrt{\frac{2s}{a} }

where a = acceleration due to gravity and s = vertical height

t = \sqrt{\frac{2 x 1.5 }{9.8} } = 0.55 s

 

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m

8 0
3 years ago
A point from which the position of other objects can be described is called what?
Alisiya [41]

Answer:

Reference Point

Explanation:

3 0
4 years ago
Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n
pickupchik [31]

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0
3 years ago
Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume
nikklg [1K]
Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.
7 0
3 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
3 years ago
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