Answer:
W= 61.3 N
Explanation:
The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:
Fg =G*ms*me / (res)²
This force, also obeys the Newton's 2nd Law, so we can write the following equation:
G*ms*me*/ (res)² = ms* a = ms*g
We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:
G*ms*me / (res)² = ms*g = W (1)
where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and
res= 4 *re = 4*6.37*10⁶ m.
Replacing all these known values in (1), we get the value of W:
W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N
Energy is the capacity of doing work
<span>The bright, visible surface of the Sun is called corona. The outermost layer of the Sun's atmosphere is called chromosphere.</span>
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
Answer:
The transmitted intensity through all polarizers is 
Explanation:
According to Malu's law the intensity of a polarized light having an initial intensity 
is mathematically represented as
Now considering the polarizer(The polarizing disk) the equation above becomes
Where n is the number of polarizers
Substituting 
for the initial intensity 3 for the n and 20° for the angle of rotation
