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seropon [69]
3 years ago
9

Why is thermal energy from the Sun transferred to Earth through electromagnetic waves instead of any other type of thermal energ

y transfer?
Physics
2 answers:
liq [111]3 years ago
8 0

Answer:

Because electromagnetic waves can travel through empty space

Explanation:

The energy that is emitted from the sun is transferred to the earth in the form of radioactive waves. These waves are originated due to the vibration between the electric and magnetic fields. As this energy reaches the earth, it warms the earth's atmosphere, resulting in the transfer of heat energy in three possible ways namely the conduction, convection, and radiation.

This electromagnetic waves do not require any matter for the transmission of energy, and can easily travel in empty space from the core of the sun to the earth and other nearby planets. Whereas other types of waves cannot travel in space, so it is transferred in the form of electromagnetic waves only.

Bond [772]3 years ago
7 0

Answer:

The electromagnetic waves can pass through empty space.

Explanation:

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An owl weighing 40N is sitting in a tree waiting to dive down to catch a mouse. If the owl's potential energy is 800 J with resp
Natali5045456 [20]

Answer:

<em>h = 20 m</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8 m/s^2.

The weight of an object of mass m is:

W = m.g

Thus, the GPE is:

U=W.h

Solving for h:

\displaystyle h=\frac{U}{W}

The weight of the owl is W=40 N and its GPE is U=800 J.

\displaystyle h=\frac{800}{40}=20

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3 years ago
What are the three psychological processes of memory?
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3 years ago
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Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that
lawyer [7]

As we know by the first law of thermodynamics

Q = \Delta U + W

here we know that

Q = heat given to the system

\Delta U = U_f - U_i

W = work done by the system

now here we can say

\Delta U = 180 - 78 = 102 J

W = 64 J

now we can say that heat will be given as

Q = 64 + 102 = 166 J

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

<em>B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy. </em>

8 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

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Answer:

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