Question:
Which of the following parasites are small parasitic arachnids which feeds on the blood of mammals, birds and reptiles?
Answer: TICK.
Explanation: A tick is a tiny woodland arachnid of the suborder ixodida. They are external parasites and they live by feeding on the blood of mammals, reptiles, birds, and amphibians.
What is energy in general?
Answer:
The ability to do work or cause change
Explanation:
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm diameter pipe in the basement under a pressure of 3.03x10^5 Pa, what will be the velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above?
The velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above are 1.18m/s and 2.51 * 10^5 Pa
The given parameters are:
Initial velocity: [tex]v_1 = 0.5ms^{-1}[/tex] Initial diameter [tex]d_1 = 4cm[/tex] Initial pressure: [tex]P_1 = 3.03 \times 10^5 Pa[/tex]
Start by calculating the radius using:
[tex]r = 0.5d[/tex]
So, we have:
[tex]r_1 = 0.5 \times 4cm[/tex]
[tex]r_1 = 2cm[/tex]
Express as meters
[tex]r_1 = 0.02m[/tex]
Next, calculate the area using:
[tex]A =\pi r^2[/tex]
So, we have:
[tex]A_1 = \pi \times 0.02^2[/tex]
[tex]A_1 = \pi \times 0.004[/tex]
[tex]A_1 = 0.004\pi[/tex]
Also from the question, we have:
[tex]d_2 = 2.6cm[/tex]
[tex]h = 5m[/tex]
Calculate the radius using:
[tex]r = 0.5d[/tex]
So, we have:
[tex]r_2 = 0.5 \times 2.6cm[/tex]
[tex]r_2 = 1.3cm[/tex]
Express as meters
[tex]r_2 = 0.013m[/tex]
The area is then calculated as:
[tex]A =\pi r^2[/tex]
So, we have:
[tex]A_2 = \pi \times 0.013^2[/tex]
[tex]A_2 = \pi \times 0.000169[/tex]
[tex]A_2 = 0.000169\pi[/tex]
The velocity is then calculated using:
[tex]A_1v_1 = A_2v_2[/tex]
Make v2 the subject
[tex]v_2 = \frac{A_1v_1}{A_2}[/tex]
So, we have:
[tex]v_2 = \frac{0.0004\pi \times 0.5}{0.000169\pi}[/tex]
[tex]v_2 = \frac{0.0004\times 0.5}{0.000169}[/tex]
[tex]v_2 = 1.18343195266[/tex]
Approximate
[tex]v_2 = 1.18[/tex]
The pressure is then calculated as follows:
[tex]P_1 + 0.5 \times density \times v_1^2 + density \times g \times h_1 = P_2 + 0.5 \times density \times v_2^2 + density \times g \times h_2[/tex]
Where:
[tex]g = 9.81ms^{-1}[/tex]
[tex]density = 1000kgm^{-3[/tex]
[tex]h_1 = 0[/tex]
So, we have:
[tex]3 \times 10^5 + 0.5 \times 1000 \times 0.5^2 + 1000 \times 9.8 \times 0 = P_2 + 0.5 \times 1000 \times 1.18^2 + 1000 \times 9.8 \times 5[/tex]
[tex]300000 + 125 + 0 = P_2 + 696.2 + 49000[/tex]
Collect like terms
[tex]300000 + 125 - 696.2 - 49000 = P_2[/tex]
[tex]250428.8 = P_2[/tex]
Rewrite as:
[tex]P_2 =250428.8[/tex]
Rewrite as:
[tex]P_2 = 2.51 \times 10^5\ Pa[/tex]
Hence, the velocity and pressure are 1.18m/s and 2.51 * 10^5 Pa
Read more about pressures and velocities at:
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The velocity in the 2.6-cm diameter pipe is 2.0 m/s and the pressure in the 2.6-cm diameter pipe is 2.73x10^5 Pa.
To solve this problem, we can use the principle of continuity and the Bernoulli equation.
Continuity equation
The continuity equation states that the mass flow rate through a pipe is constant. This means that the product of the density of the fluid, the cross-sectional area of the pipe, and the velocity of the fluid is the same at all points in the pipe.
ρ * A * v = constant
where:
ρ is the density of the fluid in kg/m³
A is the cross-sectional area of the pipe in m²
v is the velocity of the fluid in m/s
Bernoulli equation
The Bernoulli equation states that the total energy of a fluid is constant along a streamline. This total energy is made up of the fluid's pressure energy, kinetic energy, and potential energy.
P + 1/2ρv² + ρgh = constant
where:
P is the pressure of the fluid in Pa
ρ is the density of the fluid in kg/m³
v is the velocity of the fluid in m/s
g is the acceleration due to gravity in m/s²
h is the height of the fluid above a reference point in m
Solving for the velocity and pressure in the 2.6-cm diameter pipe
We can use the continuity equation to solve for the velocity in the 2.6-cm diameter pipe.
ρ * π(0.02 m)² * v_2 = ρ * π(0.04 m)² * v_1
where:
v_1 is the velocity in the 4.0-cm diameter pipe
v_2 is the velocity in the 2.6-cm diameter pipe
Substituting in the known values, we get:
v_2 = (π(0.04 m)² * v_1) / π(0.02 m)²
v_2 = 4v_1
Therefore, the velocity in the 2.6-cm diameter pipe is four times greater than the velocity in the 4.0-cm diameter pipe. Since the velocity in the 4.0-cm diameter pipe is 0.50 m/s, the velocity in the 2.6-cm diameter pipe is 2.0 m/s.
We can use the Bernoulli equation to solve for the pressure in the 2.6-cm diameter pipe.
P_1 + 1/2ρv_1² + ρgh_1 = P_2 + 1/2ρv_2² + ρgh_2
where:
P_1 is the pressure in the 4.0-cm diameter pipe
P_2 is the pressure in the 2.6-cm diameter pipe
h_1 is the height of the water in the 4.0-cm diameter pipe above a reference point
h_2 is the height of the water in the 2.6-cm diameter pipe above a reference point
Assuming that the height of the water in the two pipes is the same, we can cancel out the ρgh terms.
P_1 + 1/2ρv_1² = P_2 + 1/2ρv_2²
Substituting in the known values, we get:
3.03x10^5 Pa + 1/2ρ(0.50 m/s)² = P_2 + 1/2ρ(2.0 m/s)²
P_2 = 3.03x10^5 Pa + 1/2ρ(0.50 m/s)² - 1/2ρ(2.0 m/s)²
P_2 = 2.73x10^5 Pa
Therefore, the pressure in the 2.6-cm diameter pipe is 2.73x10^5 Pa.
For such more question on velocity
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A 2-m3 rigid tank initially contains air at 100 kPa and 22oC. The tank is connected to a supply line through a valve. Air is flowing in the supply line at 600 kPa and 22oC. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the supply line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77oC. Determine (a) the mass of air that has entered the tank and (b) the amount of heat transfer. [R=0.287 kPa.m3/kgK]
Answer:
a. 9.5838kg
b. 340.321kj
Explanation:
See attachment please
What is Absorption in your own words
Answer:
The process by which something absorbs or is absorbed by something else.
Technician A says that the first step in diagnosing engine condition is to perform a thorough visual inspection. Technician B says that oil leaks can be found by measuring straight up from where the oil drips on the floor. Who is right?
A. Technician A only
B. Technician B only
C. Both technicians A and B
D. Neither technician A nor B
Why is there a ring of satellites around the Earth’s equator roughly 22,000 miles away from the Earth?
Answer: satellite is a artificial sub planet and every sub planet must around to it's main planet for their connect.
Explanation:
Satellite mainly 3types.
1. GEO (22,000mi away)
2. MEO (10,000mi away)
3. LEO (1,000-3,000mi away)
If any satellite will stabilised more then 22,000mi then it's can't around as a ring to world. And takes long time more then 24hours. And it's periodic time may disturbance.