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Sedbober [7]
3 years ago
8

select and explain three features of the solar panel that maximises the final temperature of the water​

Physics
1 answer:
MatroZZZ [7]3 years ago
7 0

Answer:

Irradiance (sunlight intensity or power), in Watts per square meter falling on a flat surface. ... Air Mass refers to “thickness” and clarity of the air through which the sunlight passes to reach the modules (sun angle affects this value). ... Cell temperature , which will differ from ambient air temperature

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If an airplane travels a distance of 500 km in 5 hours, what is its average speed?
kolezko [41]

The plane travels 100km per hour


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Liquids and gases transfer heat by ____.
vodka [1.7K]
Fluids always transfer heat by convection. So, your answer is B. That's how a oven works actually!
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4 years ago
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The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

v_y^2 =v_0^2 +2ay

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

 v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s  v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0
4 years ago
A copper (Cu) atom has 29 protons, 34 neutrons, and 29 electrons. What is
Arturiano [62]
B. 29 atomic number is the number of protons
4 0
3 years ago
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 re
Komok [63]

Answer:

(a) α = -0.16 rad/s²

(b) t = 33.2 s

Explanation:

(a)

Applying 3rd equation of motion on the circular motion of the tire:

2αθ = ωf² - ωi²

where,

α = angular acceleration = ?

ωf = final angular velocity = 0 rad/s (tire finally stops)

ωi = initial angular velocity = 5.45 rad/s

θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad

Therefore,

2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²

α = -(29.7 rad²/s²)/(57.6π rad)

<u>α = -0.16 rad/s²</u>

<u>Negative sign shows deceleration</u>

<u></u>

(b)

Now, we apply 1st equation of motion:

ωf = ωi + αt

0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t

t = (5.45 rad/s)/(0.16 rad/s²)

<u>t = 33.2 s</u>

6 0
4 years ago
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