A large in-falling fragment could be tracked using radar. Explain how distance, speed, and the direction of motion, of the fragm
1 answer:
Answer:
speed, distance and direction of motion of the object can be determined by analyzing the radio wave.
Explanation:
We know that radar operates by transmitting radio waves to a destination and these waves are comes back to the receiver station. By Considering this transmission and receiver process, we can measure the distance, velocity and path of an object's movement.
Distance can be assessed by taking following consideration, the velocity of the waves is V. It can help to assess the time made for the waves to be emitted by the radar and felt by the receiver, let the time be t.
Therefore distance can be determine as D= v*t/2,
here 2 signifies that the distance travelled by the wave in either direction ( from transmitter to receiver and vice verse)
Using the source wave frequency, speed can be computed. In a specific frequency, the radar starts sending out the frequencies and the reflected wave will have a distinct frequency. The velocity can be determine by
whereis the change in frequency and
c is the speed of light (the wave).
Direction can be determine by applying above principle. change in frequency is used to determine the direction in the following way:
When the frequency transition is very low, the object moves away from the radar and vice verse.
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Answer:
(a) i. The minimum work required to pump the water used per day is
291.85 kJ
ii. The minimum power rating of the pump is 40.53 Watts
(b) i. The flow velocity at the house when a faucet in the house is open where the diameter of the pipe is 1.25 cm is 2.87 m/s
ii. The pressure at the well when the faucet in the house is open is
837.843 kPa.
Explanation:
We note the variables of the question as follows;
Depth of well = 35 m deep
Height of house above the top of the well = 50 m
Density of water = 1000 kg/m³
Volume of water pumped per day = 0.35 m³
Duration of pumping of water per day = 2 hours
(a) i. We note that the energy required to pump the water is equivalent to the potential energy gained by the water at the house. That is
Energy to pump water = Potential Energy = m·g·h
Where:
m = Mass of the water
g = Acceleration due to gravity
h = Height of the house above the bottom of the well
Therefore,
Mass of the water = Density of the water × Volume of water pumped
= 1000 kg/m³ × 0.35 m³ = 350 kg
Therefore P.E. = 350 × 9.81 × (50 + 35) = 291847.5 J
Work done = Energy = 291847.5 J
Minimum work required to pump the water used per day = 291847.5 J
= 291.85 kJ
ii. Power is the rate at which work is done.
Power =
Since the time available to pump the water each day is 2 hours or 7200 seconds, therefore we have
Power = 291847.5 J/ 7200 s = 40.53 J/s or 40.53 Watts
(b)
i. If the velocity in the 3.0 cm pipe is 0.5 m/s
Then we have the flow-rate as Q = v₁ ×A₁
Where:
v₁ = Velocity of flow in the 3.0 cm pipe = 0.
A₁ = Cross sectional area of 3.0 cm pipe
As the flow rate will be constant for continuity, then the flow-rate at the faucet will also be equal to Q
That is Q = 0.5 m/s × π × (0.03 m)²/4 = 3.5 × 10⁻⁴ m³/s
Therefore the velocity at the faucet will be given by
Q = v₂ × A₂
∴ v₂ = Q/A₂
Where:
v₂ = velocity at the house the where the diameter of the pipe is 1.25 cm
A₂ = Cross sectional area of 1.25 cm pipe = 1.23 × 10⁻⁴ m²
Therefore v₂ = (3.5 × 10⁻⁴ m³/s)/(1.23 × 10⁻⁴ m²) = 2.87 m/s
ii. The pressure at the well is given by Bernoulli's equation,
P₁ + 1/2·ρ·v₁² + ρ·g·h₁ = P₂ + 1/2·ρ·v₂² + ρ·g·h₂
If h₁ is taken as the reference point, then h₁ = 0 m
Also since P₂ is opened to the atmosphere, we take P₂ = 0
Therefore
P₁ + 1/2·ρ·v₁² + 0 = 0 + 1/2·ρ·v₂² + ρ·g·h₂
P₁ + 1/2·ρ·v₁² = 1/2·ρ·v₂² + ρ·g·h₂
P₁ = 1/2·ρ·v₂² + ρ·g·h₂ - 1/2·ρ·v₁²
= 1/2 × 1000 × 2.87² + 1000 × 9.81 × 85 - 1/2 × 1000 × 0.5²
= 837843.45 Pa = 837.843 kPa