Suppose we take a 1 m long uniform bar and support it at the 33 cm mark. Hanging a 0.15 kg mass on the short end of the beam res
1 answer:
Answer:
The mass of the beam is 0.074 kg
Explanation:
Given;
length of the uniform bar, = 1m = 100 cm
Set up this system with the given mass and support;
0-----------------33cm-----------------------------------100cm
↓ Δ ↓
0.15kg m
Where;
m is mass of the uniform bar
Apply the principle of moment to determine the value of "m"
sum of anticlockwise moment = sum of clockwise moment
0.15kg(33 - 0) = m(100 - 33)
0.15(33) = m(67)
Therefore, the mass of the beam is 0.074 kg
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Answer:
a)6.34 x W/m²
b)1.37 x W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x J
and the area of the sun is given by,
A= 4π => 4π
A=6.15 x m²
= 3.9 x
/ 6.15 x
=><u> 6.34 x </u>
<u>W/m²</u>
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π
Now R= 1.5 x m
A= 4π x 1.5 x =>2.83 x
m²
The power that each square meter of Earths surface receives
= 3.9 x
/2.83 x
=><u>1.37 x </u>
<u> W/m²</u>
<u />
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.