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andreev551 [17]
3 years ago
14

What method of heat transfer does heat energy use to reach earth from the sun?

Physics
2 answers:
Alchen [17]3 years ago
4 0
Through radiation of the sun's rays. 
quester [9]3 years ago
4 0

Answer: Radiation

Explanation:

Heat transfer through radiation. Radiation is one of the mode of heat transfer and is the transfer of heat from one body to another without passing through any intervening medium. It is the heat energy directly from the sun. This energy reaches the earth directly without obstruction.

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A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
2 years ago
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Alexxandr [17]
Number 4 I think
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2 years ago
A 60 kilogram student jumps down from a laboratory counter. At the instant he lands on the floor hus speed is 3 meters per secon
erastovalidia [21]

As per Newton's law rate of change in momentum is net force

so we can write it as

F = \frac{dP}{dt}

F = \frac{m(v_f - v_i)}{\Delta t}

now we know that

m = 60 kg

v_f = 3 m/s

v_i = 0

\Delta t= 0.2 s

from above equation

F = \frac{60(3 - 0)}{0.2} = 900 N

so he will experience 900 N force in above case

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C. Kevlar

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Explanation:

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