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Svetradugi [14.3K]
3 years ago
5

The average intensity of sunlight at the top of the earth's atmosphere in 1390 w/m2. what is the maximum energy that a 34-m x 46

-m solar panel could collect in 2.0 hours in this sunlight
Physics
1 answer:
slavikrds [6]3 years ago
7 0
Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy
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The mass of an electron is...
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but the most complex atom has only around 100 of them.</span>
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3 years ago
How can we realize that light travel in straight line ?​
Norma-Jean [14]

Answer:

It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.

Explanation:

The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.

One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.  

3 0
3 years ago
How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have
cricket20 [7]
1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J
So the power output is 
P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.
6 0
3 years ago
Read 2 more answers
Explanation for question i &amp; ii. Thank you.
suter [353]
Do u know Chinese ?if yes I can explain to u easily
6 0
3 years ago
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
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