Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
Light waves can be from any color, depends on what it is bouncing no or reflecting off of.
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
<h3>
Time spent in air by the baseball</h3>
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
<h3>Horizontal distance traveled by the baseball</h3>
R = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Learn more about horizontal distance here: brainly.com/question/24784992
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Answer:
1.57 s
Explanation:
Since the motion of the hammer is a uniformly accelerated motion, the distance covered by the hammer in a time t is
Where, in this case
S = 2.0 m is the distance covered
a = 1.62 m/s^2 is the acceleration due to gravity
t is the time taken
Re-arranging the equation, we can find the time the hammer takes:
