Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t
1 answer:
This question is incomplete, the complete question is;
The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters
a) Find the volume V in terms of x and L
b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.
Answer:
a) the volume V in terms of x and L is ((√3/4)x²L) m³
b) required expression is (2/(3)^(1/u))√(At/L)
Explanation:
Given that;
form the question and image below;
triangular prism ends are equilateral triangle
side length = x meter
Dimension of the prism = L meter
Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter
Volume of the triangular prism = Area × height
= √3/4 (x)² × L
V = ((√3/4)x²L) m³
Therefore, the volume V in terms of x and L is ((√3/4)x²L) m³
b)
Rate of water entering = A m³/hr
Depth of water tank = d meter
Time = t
Length of prism = L
now Rate of water entering is A m³/hr
dv/d = A [ V = ((√3/4)x²L) m³ ]
and
dv/dt = √3/4 [2x dx/dt ] L { L is constant }
so
A = √3/4 [2x dx/dt ] L
∫A dt = √3/2 [ Lx dx ] { Integrate both sides}
At = √3/2 × Lx × x²/2
x² = uAt / √3L { we find square root of both sides}
x = √( uAt / √3L )
x = (2/(3)^(1/u))√(At/L)
Therefore; required expression is (2/(3)^(1/u))√(At/L)
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Answer:
The total distance is 381.5 [m]
Explanation:
In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
a = acceleration = 2 [m/s²]
t = time = 7 [s]
Vf = 0 + (2*7)
Vf = 14 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
14² = 0 + (2*7*x)
x = 196/(14)
x = 14 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.
We must calculate the final velocity.
Vf = final velocity [m/s]
Vi = initial velocity = 14 [m/s]
a = desacceleration = 4 [m/s²]
t = time = 8 [s]
Vf = 24 + (4*8)
Vf = 56 [m/s]
With this velocity, we can calculate the displacement using the following expression.
where
x = distance traveled [m]
56² = 14² + (2*4*x)
x = 2940/(8)
x = 367.5 [m]
Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.
Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
