Answer : The volume of water in the pool is, 2473 ft³
Explanation :
First we have to calculate the average depth.
Given:
Diameter = 30 ft
Depth range : 1 to 6 ft linearly
average depth =
Now we have to calculate the volume of water in the pool.
Volume = area × average depth
V = π × (radius)² × 3.5
V = π × (30/2)² × 3.5
V = π × (15)² × 3.5
V = π × 225 × 3.5
V = 3.14 × 225 × 3.5
V = 2472.75 ft³ ≈ 2473 ft³
Therefore, the volume of water in the pool is, 2473 ft³
Answer:
The distance close to the sidewalk can the flower pot fall is
x = 14.83 m
Explanation:
Given
Δt = 0.300 s
d = 19.6 m
h = 1.79 m
Knowing as the velocity of the sound as a 330 m/s
t = (19.6 - 1.79)m / 330 m/s
t = 0.0539 s
Total time
tₙ = 0.3 + 0.0539 = 0.3539 s
Time for flower-pot
s = ¹/₂ * g * t²
tₐ = √[(2 * 17.81m)/9.81m/s²]
tₐ = 1.34 s, t' = 0.3539
1.34 - 0.3539 = 0.9861 s
19.6 m - x = ¹/₂* g * t ²
x = 19.6 - ¹/₂ * (9.81) * (0.9861)²
x = 14.83 m
The thermal efficiency of an engine is
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work
In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
Answer: 0.6
Explanation:
If we draw a free body diagram of the box we will have the following:
Net force in the x-axis:
(1)
Net force in the y-axis:
(2)
Where:
is the normal force
is the weight of the box
is the force exerted on the box
is the angle below the horizontal
is the friction force, being the coefficient of kinetic friction
Isolating from (1):
(3)
Substituting (3) in (2):
(4)
Finding :
(5)
(6)
Finally:
(5)
Answer:
Explanation:
a ) If x be the position of n the bright fringe on the screen , following formula holds .
x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .
If we increase the value of λ or wave length, x will increase so central fringe along with all the fringes will shift away from the centre .
If we increase the value of D or screen distance , it will also increase x , so fringes along with central fringe will shift away from the center.
b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .
So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .