Answer:
5.4 × 10⁸ W/m²
Explanation:
Given that:
The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W
the mass of the exoplanet = 5.972 × 10²⁴ kg
radius of the earth = 1.27 × 10⁷ m
half the distance (i.e radius r ) = 7.5 × 10¹⁰ m
a) What is the intensity of Betelgeuse at the "earth’s" surface?
The Intensity of Betelgeuse can be determined by using the formula:


I = 544097698.8 W/m²
I = 5.4 × 10⁸ W/m²
K:potassium
Cu:copper
Cl: chlorine
Answer:N=0
Explanation:
Given


both blocks experiencing free fall so net weight of block during free fall is zero thus there is no normal reaction between them.
N=0
Answer:
<em>0.25</em>
Explanation:
According to newtons law of motion
\sum F_x = ma
F_f = ma
nR = ma
nmg = ma
ng = a
n = a/g
g is the acceleration due to gravity
Given
a = 2.42m/s²
g = 9.8m/s²
Substitute into the formula;
n = 2.42/9.8
n = 0.25
<em>Hence the coefficient of kinetic friction is 0.25</em>
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