That's "<em><u>insolation</u></em>" ... not "insulation".
'Insolation' is simply the intensity of solar radiation over some area.
If 200 kW of radiation is shining on 300 m² of area, then the insolation is
(200 kW) / (300 m²) = <em>(666 and 2/3) watt/m²</em> .
Note that this is the intensity of the <em><u>incident</u></em> radiation. It doesn't say anything
about how much soaks in or how much bounces off.
Wait !
I just looked back at the choices, and realized that I didn't answer the question
at all. I have no idea what "1 sun" means. Forgive me. I have stolen your
points, and I am filled with remorse.
Wait again !
I found it, through literally several seconds of online research.
1 sun = 1 kW/m².
So 2/3 of a kW per m² = 2/3 of 1 sun
That's between 0.5 sun and 1.0 sun.
I feel better now, and plus, I learned something.
Answer:
The electric field value is 240 N/C
Explanation:
Given that,
Distance = 5.0 mm
Potential difference = 1.2 V
We need to calculate the electric field value
Using formula of potential difference
Where, E = electric field
V = potential difference
d = distance
Put the value into the formula
Hence, The electric field value is 240 N/C
The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
Learn more about velocity here: brainly.com/question/4931057
#SPJ1
Answer:
he tail of the arrow moves a distance of 0.5 m as the arrow is shot. yare yare daze
Explanation:
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
