Answer:
The process by which something absorbs or is absorbed by something else.
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.
What is energy in general?
Answer:
The ability to do work or cause change
Explanation:
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
3. Vynich of the following parasites are small parasitic arachnids which leeu
blood of mammals, birds and reptiles?
A. Roundworms
B. Whipworms
C. Ticks
D. Ear mites
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.
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
What are three common hazardous
conditions that weather causes in
Pennsylvania, and how do they make
driving conditions hazardous?
Answer this question in at least 3 complete sentences.
Answer:
SNOW SQUALLS - Squalls are substantial snow showers that cause white-out conditions. A couple of squalls are conceivable at the beginning of today through mid-evening. The most elevated possibility is north of the expressway. COLD - Because of away from and a new cover of day off, have plunged to the youngsters and 20s. There are frigid untreated surfaces. WINDS – One can expect supported breezes 15-25 mph this evening, with blasts to 40 mph. Solid breezes, alongside temperatures during the 30s, will create wind chills during the 20s. FOG - Perceivability is improving, however parts of Lancaster, Lebanon, Schuylkill, Berks, and Dauphin Provinces have decreased perceivability. With temperatures beneath freezing, mist can stick to the streets and produce frigid conditions.Explanation:
An effect day in Pennsylvania is a day that highlights climate that will probably upset your typical every day calendar or schedule. At the point when the News 8 Tempest Group accept climate arrives at this limit, a yellow triangle with a shout imprint will show up on the figure demonstrating the climate will cause an effect. You may likewise observe the symbol utilized during a day-part figure when the News 8 Tempest Group accepts certain times may have a greater amount of an effect than others. An effect day's climate isn't viewed as outrageous or serious, perilous climate.
Light downpour, day off, ice during a morning or night drive would be instances of effect climate. Most Pennsylvania workers can deal with driving in light downpour, day off, ice, however light downpour, day off, ice at the pinnacle of busy time can upset travel or cause mishaps because of the high volume of drivers. Thick mist over the Valley which could likewise modify a drive would be delegated an effect day.
Moderate cold throughout the fall, winter, spring months, or warmth in the mid year months that may constrain open air presentation to the components would get this assignment.
The Susquehanna Valley is a significant agrarian community for the locale. Pre-summer or late-summer ice or freezes that could harm or murder vegetation would fit into this class.
Days that component consume bans from delayed droughts combined with solid breeze blasts likewise would be considered to conceivably make an effect a day ahead.
Name 2 cellular structures (organelles) that plant cells have, but animal cells don’t.
Then, choose one and explain why an animal cell does not need it.
Answer:
a) -Cell Wall
- Plastids such as Chloroplasts
b) Please refer to Explanation
Explanation:
Organelle structure and components are mostly similar with both plant and animal cells, but in certain areas the differ, because they do different things to survive.
One of the ways they differ is their organelle content. There are some organalles that Plant cells have that Animal cells do not.
Two examples of them are;
- Cell Wall
- Plastids such as Chloroplasts
b) Cell Wall - Why does an Animal cell not need it?
Animal cells do not need cell walls for one reason, bones or rather a lack thereof.
Cell walls act as a sort of Exoskeleton for cells as they maintain the cell's shape. This leads to a rigidity that enables the plant to stand up on its own without the need for bones like in animals.
Why is it important to collect forensic evidence and witness accounts as soon as possible after an accident?
Answer:
If you have enough forensic evidence than you can be able to figure out who committed the accident or the burglary
Explanation:
Please give BRAINLIEST ANSWER └[T‸T]┘
When considering what you want your life to be like in the future, what are some helpful questions to ask yourself? Check all that apply. Where do I want to live? How many teachers have I had? What activities do I enjoy? What personal goals will I have accomplished? What things would my job need to include in order to make me feel satisfied? What are the weaknesses of my family members?
Answer:
ACDE
Explanation:
ACDE
The possible questions based on text will be:
Where do I want to live?What activities do I enjoy?What personal goals will I have accomplished?What things would my job need to include in order to make me feel satisfied?What one wants to consider?When one wants to consider where to live in the future, one should consider where he or she wants to live. Also, the person should know th activities that they enjoy and the personal goals that they've achieved and still want to achieve. The number of teachers that the person has have and the weakness of one family members aren't determinant.
A meaningful life may be defined as a life in which one feels engaged with things, connected to purpose, and capable to connect your gifts and passions with your highest values. This form of life is a construct that has to do with the purpose, significance, fulfillment, and satisfaction of life.
Therefore, Where do I want to live?
What activities do I enjoy?are the possible questions.
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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
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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.
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