Answer:
D............................
Answer:
The diameter of the moon's image is 0.31 cm.
Explanation:
Given that,
Focal length = 34.3 cm
Diameter of the moon
Mean distance from the earth
At that distance the object is assumes to be at infinity. hence the image will be formed at a distance equal to focal length
So, the image distance is 34.3 cm.
We need to calculate the diameter of the moon's image
Using formula of magnification
Hence, The diameter of the moon's image is 0.31 cm.
Well, all the words are common, but I try to answer in my way. The way how light is used to study space is finding distance between the planets, which is called a light year. According to the <span>scientist's beliefs the space is but a vaccum and the only body that can travel through it is light, that's why light is very importand for discovering space.
</span>Hope you will find it helpful!
Answer:
a) 269.23 N
b) Fr = 70.77 N
c) Work output = 5250 J
d) Work Input = 6630 J
e) Mechanical Advantage
- ideal mechanical advantage = 3.9
- Actual Mechanical Advantage = 3.09
Explanation:
a) In an ideal machine, all the work input is converted to work output.
Work output = F₀ × d₀ = 1050 × 5 = 5250 J
Work input = Fᵢ × dᵢ = 19.5 Fᵢ
19.5 Fᵢ = 5250
Fᵢ = 269.23 N
b) Real effort = 340 N
Since 269.23 N is the force required to lift the 1050 N load in this pulley system, the rest of the 340 N force services frictional forces.
Frictional force = 340 - 269.23 = 70.77 N
c) Output work = Work done by the load = F₀ × d₀ = 1050 × 5 = 5250 J
d) Input Work = Work done by the Effort = Fᵢ × dᵢ = 340 × 19.5 = 6630 J
e) Ideal Mechanical advantage = distance travelled by effort/distance travelled by load
ideal mechanical advantage = 19.5/5 = 3.9
Actual Mechanical Advantage = Load/Effort = 1050/340 = 3.09
Efficiency = Work output/Work input = 5250/6630 = 0.792