Answer:
A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)
a. 45%
b. 29%
c. 71%
d. 50%
The correct answer is d.
d. 50%
Explanation:
Fan cart acceleration = 1.6 m/s²
Thrust = 0.25×π×D²×ρ×v×Δv
where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D
or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv
=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)
That is the thrust reduces by 50 %
Answer:
they cross over one another between charge.
Answer: B) 0.00337 m3.
Explanation:
Given data:
Mass of the ball = 10kg
Weight of the ball in air = 98N
Weight of the ball in water = 65N
Solution:
To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.
= 98 - 65 / 9.8
= 33 / 9.8
= 3.37kg
The volume of the ball is 3.37kg
The density of water is 1kg per Liter.
So 3.37 kg of water would have a volume of 3.37 Liters.
Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).
