The final expression for the charge Q(t) at any time t is given as:Q(t) = CV(t) = 2.5 × 10^-11 e- t/RC
To find the charge on the capacitor at any time t, we need to find the total current in the circuit and then find the charge using the formula Q = CV, where V is the potential difference across the capacitor.Let's find the total current in the circuit using the formula:
I = (1/LC)½ x (e- Rt/2L) sin(wt - φ)
where, L = inductance C = capacitance R = resistance ω = (1/LC)½ = 5000 sinφ = RωL = 260 × 5000 × 0.2 = 2600
Let's now substitute the given values into the formula and simplify:I = (1/(0.2 × 1.25 × 10^-5))½ x (e- 260t/2 × 0.2) sin(5000t - φ)I = 10^5 x (e- 130t) sin(5000t - φ). Let's now find the charge Q on the capacitor using the formula:
Q = CV where, C = capacitance V = potential difference across the capacitor. To find the potential difference across the capacitor, we need to find the current passing through it, which is given as the total current minus the current passing through the inductor. Let's find the current passing through the inductor using the formula:
I L = I x sin(wt - φ)IL = I x sin(5000t - φ).The potential difference across the capacitor can be calculated using the formula:V C = V 0 × e- t/RC where, V0 = initial potential difference across the capacitor R = resistance of the circuit C = capacitance of the circuit. Let's now find the current passing through the capacitor:I C = (I - I L)I C = I - I L
Now we have all the necessary formulas to find the charge Q(t) at any time t. Let's substitute the given values into the formulas and simplify:
I = 10^5 x (e- 130t) sin(5000t - φ)IL = I x sin(5000t - φ)IC = I - I LVC = V0 × e- t/RCQ = CVCI = I - I L = 10^5 x (e- 130t) sin(5000t - φ) - I sin(5000t - φ)V C = V 0 × e- t/RC = 2 × 10^-6 e- t/RCQ = C × V C = (1.25 × 10^-5) × (2 × 10^-6) e- t/RC = 2.5 × 10^-11 e- t/RC
Now, let's substitute the values of I and V C into the formula for IC to obtain:IC = 10^5 × (e- 130t) sin(5000t - φ) - 10^5 sin(5000t - φ) × e- t/RC. Therefore final expression for the charge Q(t) at any time t is given as:Q(t) = CV(t) = 2.5 × 10^-11 e- t/RC
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We can use the equation [tex]q(t) = C.V(t)[/tex] to calculate the charge q (t) on the capacitor at any time t: [tex]q(t) = 1,25 . 10-5 Farad.V(t)[/tex].
The charge on a capacitor in a series circuit at any time t is given by the equation [tex]q(t) = C.V(t)[/tex], where C is the capacitance of the capacitor and V(t) is the voltage across the capacitor at time t.
In the given circuit, the capacitance of the capacitor is 1.25 x 10-5 Farad, and the initial charge on the capacitor is 2 x 10-6 Coulomb. Therefore, to find the charge q(t) on the capacitor at any time t, we need to find the voltage V(t) across the capacitor at time t.
To do this, we must first calculate the total inductance and resistance in the circuit. The total inductance is the sum of the inductances of each inductor, so the total inductance in this circuit is 0.2 Henry. The total resistance is the sum of the resistances of each resistor, so the total resistance in this circuit is 260 Ohms.
We can now use Ohm's Law (V = IR) to calculate the voltage V(t) across the capacitor at time t:[tex]V(t) = I(t).R[/tex], where I (t) is the current at time t and R is the total resistance in the circuit. Since the inductance of the circuit is 0.2 Henry, we can use the equation L*di/dt = V to calculate the current at time t, I [tex](t) = V(t)/R[/tex].
Substituting this into Ohm's Law, we get: V(t) = (V(t)/R)*R. Solving for V(t), we get V(t) = V(t). Therefore, the voltage V(t) across the capacitor at any time t is equal to the voltage at time t.
Finally, we can use the equation [tex]q(t) = C.V(t)[/tex]to calculate the charge q(t) on the capacitor at any time t: [tex]q(t) = 1,25 . 10-5 Farad.V(t)[/tex].
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as a 4.4-kg object moves from (2 i 5 j) m to (6 i - 2 j) m, the constant resultant force acting on it is equal to (4 i - 3 j) n. if the speed of the object at the initial position is 4.9 m/s, what is the work done by the force, and what is its kinetic energy at its final position? as your answer in canvas, write the kinetic energy in joules.\
The kinetic energy of the object at its final position is 90.98 J.Given,Mass, m = 4.4 kg Initial position, r1 = (2 i + 5 j) m, Final position, r2 = (6 i − 2 j) m ,Initial velocity, u = 4.9 m/s ,Constant resultant force, F = (4 i − 3 j) N .To find the work done by the force,First, we need to find the displacement vector = r2 - r1= (6 i − 2 j) - (2 i + 5 j)= (6 - 2) i + (-2 - 5) j= 4 i - 7 j
Magnitude of the displacement vector,= √(4² + (-7)²)= √65 m Now, we can find the work done by the force,W = F.s= (4 i - 3 j) . (4 i - 7 j)= 4(4) + 3(7)= 37 J
Therefore, the work done by the force is 37 J.
To find the kinetic energy of the object at its final position,First, we need to find the final velocity of the object by using the work-energy principle.Initial kinetic energy, K1 = (1/2)mu²= (1/2) × 4.4 × (4.9)²= 53.98 J
Work done by the force, W = 37 JFinal kinetic energy, K2 = K1 + W= 53.98 + 37= 90.98 JTherefore, the kinetic energy of the object at its final position is 90.98 J.
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what are two characteristics of net forces that are balanced
Balanced net forces have equal and opposing forces that cancel each other out and provide a net force of zero, which does not alter the motion of an item.
An object's velocity remains constant and motion is unaltered when the net forces acting on it are balanced. This indicates that the thing is either stationary or moving continuously. When the forces exerted on an item are opposing in direction and of equal magnitude, they are said to be balanced forces. The forces in this situation cancel one another out, leaving a net force of zero. This can happen when one force is applied to an item and that object applies an equal and opposite force in the opposite direction to another object. It can also happen when two or more forces are applied in opposing directions and of equal magnitude. Understanding equilibrium and stability in physics requires a knowledge of the idea of balanced forces.
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when the air is in geostrophic balance, it flows _______ to isobars _________ the boundary layer.
When the air is in geostrophic balance, it flows parallel to isobars and above the boundary layer.
Geostrophic balance is a state of balance between the pressure gradient force and the Coriolis force, where the pressure gradient force is directed from higher to lower pressure, and the Coriolis force is perpendicular to the direction of motion. In this state, the wind flows parallel to the isobars, with the pressure gradient force and the Coriolis force balancing each other out.
The boundary layer is the layer of air near the Earth's surface where friction between the air and the surface slows down the wind and causes it to flow in a more complex manner, with the wind direction changing with height. However, in the geostrophic flow regime, the wind flow is typically above the boundary layer and thus not affected by surface friction.
Therefore, in geostrophic balance, the wind flows parallel to isobars and above the boundary layer, with little to no effect from the Earth's surface.
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A bus engine transfers chemical potential energy into ___ so that the bus moves.
a. kinetic energy
b. thermal energy
c. gravitational potential energy
d. electrical energy
an electromagnetic wave is transporting energy in the positive y direction. at one point and one instant the magnetic field is in the positive x direction. the electric field at that point and instant points in the
Energy is being transported in the positive y direction by an electromagnetic wave. The magnetic field is in the positive x direction at one spot and one moment. At that precise moment, the electric field is oriented in the "negative z" direction.
The given electromagnetic wave is transporting energy in the positive y direction. At one point and one instant, the magnetic field is in the positive x direction. Now we have to find the direction of the electric field at that point and instant. According to the right-hand rule, when the magnetic field is directed towards the positive x-axis, the electric field will be directed downwards along the negative z-axis. Therefore, the electric field at that point and instant points in the negative z direction.
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Determine the power of water transferred each hour through the dam. 500 m² are cleared every hour. The height of the dam is 500m.
Answer:
The power of water transferred each hour through a 500m high dam if 500m² are cleared every hour is approximately 4.41 GW
Explanation:
To answer this question, we need to know the density of water, the gravitational acceleration, and the efficiency of the dam. Let's assume that the density of water is 1000 kg/m³, the gravitational acceleration is 9.81 m/s², and the efficiency of the dam is 100%.
The power of water transferred each hour through the dam is given by the formula:
Power = Flow rate x Density x Gravity x Height x Efficiency
where Flow rate is the volume of water that passes through the dam each second, Density is the density of water, Gravity is the gravitational acceleration, Height is the height of the dam, and Efficiency is the efficiency of the dam.
First, let's calculate the flow rate:
Flow rate = Area x Velocity
where Area is the cleared area of 500m² and Velocity is the speed of water passing through the dam.
Assuming that the water is moving at a constant speed, we can use the formula:
Velocity = Height / Time
where Time is the time it takes for the water to pass through the dam.
Since the height of the dam is 500m and we want to know the power transferred each hour, we can convert the time to seconds as follows:
Time = 1 hour / 3600 seconds per hour = 0.000277778 hours
So, the velocity of the water is:
Velocity = 500m / 0.000277778 hours = 1,800,000 m/s
Now we can calculate the flow rate:
Flow rate = 500m² x 1,800,000 m/s = 900,000 m³/s
Finally, we can calculate the power of water transferred each hour through the dam:
Power = Flow rate x Density x Gravity x Height x Efficiency
Power = 900,000 m³/s x 1000 kg/m³ x 9.81 m/s² x 500m x 1
Power = 4,405,500,000 watts or approximately 4.41 GW
Therefore, the power of water transferred each hour through a 500m high dam if 500m² are cleared every hour is approximately 4.41 GW.
If I heated up a glass of 100 grams of water, and the temperature changed from 25℃ to 31℃, how much heat was needed to do that (in calories)?
Answer:
6° because some heat is released out of surrounding. if 100 over six which is equal to sixtenn point four
When the rock hlt Cesar, the impact was softened by several protective features of the head. Which of the following structures would have helped to protect the brain from the external force? View Available Hint() Bone Oligodendrocytes Cerebrospinal fluid Basal ganglia Hair Dura mater White matter
The structure that would have helped to protect the brain from the external force when the rock hit Cesar are as follows: Dura mater and Cerebrospinal fluid.
What is the central nervous system? The central nervous system (CNS) is responsible for processing incoming stimuli from the peripheral nervous system and producing a coordinated response. It includes the brain and the spinal cord.
The brain is the largest component of the CNS, comprising 2% of the body's weight but consuming about 20% of its oxygen and nutrients. It consists of three main parts: the brainstem, the cerebellum, and the cerebrum.
The brainstem is responsible for regulating critical functions like respiration, circulation, and digestion; the cerebellum controls motor coordination, and the cerebrum is the area of the brain responsible for sensory perception, emotion, and movement.
What is external force? External forces, also known as contact forces, are forces that act on an object as a result of its interaction with its surroundings. Forces that do not require contact to take effect, such as gravitational and magnetic forces, are not considered external forces.
Examples of external forces are gravity, air resistance, tension, and friction. Dura mater and Cerebrospinal fluid as the structure that would have helped to protect the brain from the external force when the rock hit Cesar. When a rock hits Cesar, the external force created by it must be transferred to the skull, and ultimately the brain.
However, several protective features of the head help to reduce the severity of the impact. The brain is protected by two main structures: the dura mater and the cerebrospinal fluid.
The dura mater is the outermost layer of the meninges, which is a protective membrane covering the brain and spinal cord. It acts as a cushion, absorbing some of the external force generated by the impact.
Cerebrospinal fluid is a clear liquid that flows throughout the central nervous system, filling the space between the brain and the skull. It acts as a shock absorber, reducing the impact's intensity by distributing the force more evenly.
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hydroelectric dams generate electricity by question 20 options: a. using the energy of the river to produce steam. b. using run-of-the-river systems, in which turbines are placed into the natural water flow. c. water impoundment, in which dam operators control the rate of water flow to turbines. d. using generators that are placed on the bottom of a river. e. converting the kinetic energy of the water impounded behind a dam into potential energy.
Hydroelectric dams generate electricity through water impoundment, in which dam operators control the rate of water flow to turbines.
c is the correct option.
Hydroelectric dams are dams used to produce electricity. The movement of water drives turbines, which power generators that generate electricity.
The movement of water, generated by gravity, is what drives turbines. Hydroelectric dams are the most widely used renewable energy source, accounting for approximately 16% of global electricity production.
Hydroelectric dams generate electricity through water impoundment, in which dam operators control the rate of water flow to turbines.
This is the process of using turbines that are powered by the movement of water that has been dammed to generate electricity.
Turbines are powered by water that has been dammed to generate electricity, which is then sent to a power station to be used.
The electricity generated from hydroelectric dams is clean and safe, making it an important part of the renewable energy mix. They are also an essential part of the global infrastructure because they provide reliable, low-cost power.
They also assist in the management of rivers, flood control, and irrigation systems in various parts of the world.
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A ball rolls across the floor, slowing down with constant acceleration of magnitude . The ball has positive velocity ???? after rolling a distance x across the floor.
Calculate the ball's initial speed ????0 if ????= 4.51 m/s2, ????=11.17 m/s, and x=2.66 m.
A ball rolls across the floor, slowing down with a constant acceleration of magnitude a = 4.51 m/s2.
The ball has positive velocity v after rolling a distance x = 2.66 m across the floor.
To calculate the ball's initial speed v0 if
v = 11.17 m/s.
The initial velocity of the ball, v0 =?
The final velocity of the ball, v = 11.17 m/s
The acceleration of magnitude a = 4.51 m/s2
Distance travelled, x = 2.66 m
If an object has initial velocity v0, constant acceleration a, and travelled distance x, then its final velocity is given by:
v2 = v0² + 2ax
Here, the ball's initial velocity is v0, and its final velocity is v.
After substituting the given values, we have:
v2 = v0² + 2ax
=> (11.17)²
= v0² + 2(4.51)(2.66)
=> 124.57
= v0² + 25.39
=> v0² = 124.57 - 25.39
=> v0² = 99.18 => v0 = √99.18
=> v0 = 9.96 m/s
Hence, the initial velocity of the ball is v0 = 9.96 m/s.
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how many types of classifications are there for a lunar eclipse?
There are three types of lunar eclipses: total, partial, and penumbral.
During a total lunar eclipse, the moon is completely shadowed by the Earth, resulting in a reddish-brown color. In a partial lunar eclipse, only a portion of the moon is shadowed, while in a penumbral lunar eclipse, the moon passes through the Earth's outer shadow, resulting in a subtle darkening of the moon's surface. These classifications are based on the degree to which the moon passes through the Earth's shadow during the eclipse.
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What happens when thermal energy is removed from a substance?(1 point) Responses All substances will freeze. All substances will freeze. The substance loses potential energy. The substance loses potential energy. The substance’s atoms lose kinetic energy. The substance’s atoms lose kinetic energy. The substance’s particles speed up
the mass of a particular bar of gold-pressed latinum (from star trek) has a mass of 100 grams. what is the mass of this bar when it is brought to the moon?
The mass of a particular bar of gold-pressed latinum on the moon is 100 grams.
Gold-pressed latinum (GPL) is a kind of currency in the Star Trek world. Latinum, a rare silver-colored liquid, is pressed between gold layers to make GPL, which is valued in the Federation as a rare and valuable resource. The value of GPL is measured in amounts of gold. It can be used in various types of exchange and trade.
The mass of a particular bar of gold-pressed latinum when it is brought to the moon is the same as its mass on Earth. The bar's mass will stay the same no matter where it is located because mass is a constant property of an object. Mass is a measure of an object's resistance to acceleration in response to a force. It is a measure of how much matter is contained in an object.
As a result, if an object has a mass of 100 grams on Earth, it will have the same mass on the moon or any other location in the universe. Therefore, the mass of a particular bar of gold-pressed latinum when it is brought to the moon is 100 grams.
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A 12100 kg railroad car is coasting on a level, frictionless track at a speed of 19.0 m/s when a 4790 kg load is dropped onto it.
If the load is initially at rest, find the new speed of the car and the % change of the kinetic energy.
Hint 1: If the load is dropped into the car, it is like the car is "colliding� with a stationary load. If the load is stuck in the car, can they have different final velocities from one another?
The percent change in the kinetic energy of the system is [(0.5*(12100 + 4790)*1722) - 5.58 x 106] / (5.58 x 106) x 100% = 4.41%.
The 12100 kg railroad car is initially travelling at a speed of 19.0 m/s and has a kinetic energy of KE = 0.5*12100*1902 = 5.58 x 106 Joules. The 4790 kg load is dropped onto the car from rest, so its initial kinetic energy is 0.
When the load is dropped onto the car, the two objects collide and their velocities after the collision will be equal. Therefore, the final speed of both the railroad car and the load will be v = (12100*19 + 4790*0) / (12100 + 4790) = 17.2 m/s. The percent change in the kinetic energy of the system is [(0.5*(12100 + 4790)*1722) - 5.58 x 106] / (5.58 x 106) x 100% = 4.41%.
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suppose that one particle of the flow approaches a plate of a capacitor. explain what happens to the other plate of the capacitor?
The other plate of the capacitor is induced with an opposite charge through electrostatic induction as the particle of the flow approaches one plate.
As the particle of the flow approaches one plate of the capacitor, it induces an opposite charge on the other plate of the capacitor through the process of electrostatic induction. The electric field produced by the charge on the approaching plate pushes the electrons on the other plate away from the approaching plate, resulting in an accumulation of charge of the opposite sign on the other plate.
This process continues until the potential difference between the plates becomes large enough to produce a discharge, after which the process of electrostatic induction ceases. The discharge may occur in the form of a spark or a breakdown of the dielectric material separating the plates, depending on the strength of the electric field and the dielectric strength of the material.
Overall, the other plate of the capacitor experiences a temporary polarization and a buildup of charge of the opposite sign due to the approaching particle.
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A dragster is travelling east when the parachute opens and slows the dragster for 4.5 seconds at a rate of 10 m/s2 west. What was the dragster's change in velocity due to the parachute?
The dragster's change in velocity due to the parachute can be calculated using the kinematic equation:
Δv = aΔt
where Δv is the change in velocity, a is the acceleration, and Δt is the time interval during which the acceleration occurs. In this case, the dragster is initially travelling east, so its velocity is positive, and the parachute applies a force in the opposite direction, resulting in a negative acceleration.
Given that the acceleration is -10 m/s² (westward) and the time interval is 4.5 seconds, we can calculate the change in velocity as:
Δv = (-10 m/s²) x (4.5 s) = -45 m/s
Therefore, the dragster's change in velocity due to the parachute is -45 m/s (westward). This means that the dragster's velocity is reduced by 45 m/s in the westward direction over the 4.5-second interval during which the parachute is deployed.
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The change in velocity due to the parachute is -45 m/s east
What is velocity ?
Velocity is a vector quantity that describes the speed and direction of motion of an object. In other words, velocity is the rate at which an object changes its position in a specific direction.
Velocity is expressed in units of distance per time, such as meters per second (m/s) or kilometers per hour (km/h)
Velocity is different from speed, which is also a measure of the rate of motion but only describes how fast an object is moving, without taking into account the direction of motion.
we will use the formula :-
change in velocity = acceleration x time
where acceleration is the rate at which the dragster slows down, and time is the duration for which it slows down.
Here, the dragster is travelling east, and the parachute applies a force in the opposite direction (west), causing it to slow down. So, the acceleration is -10 m/s^2 (negative because it's in the opposite direction to the velocity).
The time for which the dragster slows down is 4.5 seconds.
Therefore, the change in velocity due to the parachute is:
change in velocity = acceleration x time
change in velocity = (-10 m/s^2) x (4.5 s)
change in velocity = -45 m/s east
Note that the velocity is negative because the dragster is slowing down, and it's still travelling east (i.e., in the positive direction).
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Critically discuss why the environment in most communities continue to be dirty amidst the existence of local government structures
Explanation:
# Unmanaged population distribution
# lack of sanitation programs
# lack of awareness programs
# lack of implementation of policies and rules
# carelessness of people and government
# Unmanaged waste disposal
an airplane flies due west at an airspeed of 425 mph. the wind is blowing from the northeast at 40 mph. what is the ground speed of the airplane? what is the bearing of the airplane?
An airplane flies due west at an airspeed of 425 mph and the wind is blowing from the northeast at 40 mph, the ground speed of the airplane is 385 mph, and the bearing of the airplane is 285°.
We can use the equation
GS = AS + (Wind x cos(Θ)),
Where GS is the ground speed, AS is the airspeed, and Θ is the angle between the wind and the heading of the airplane. the airspeed is 425 mph, the wind is blowing from the northeast at 40 mph, and the heading of the airplane is due west. The angle Θ is 90°. Plugging these values into the equation, we get
GS = 425 + (40 x cos(90°)) = 385 mph.
To calculate the bearing of the airplane, we can use the equation
Bearing = 180° - (Θ + (Wind ÷ AS) x 180°).
Θ is 90°, the wind is 40 mph, and the airspeed is 425 mph.
Plugging these values into the equation, we get
Bearing = 180° - (90° + (40 ÷ 425) x 180°) = 285°.
Hence , airplane flies due west at an airspeed of 425 mph and the wind is blowing from the northeast at 40 mph, the ground speed of the airplane is 385 mph, and the bearing of the airplane is 285°.
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A survey was conducted at local colleges around Madison, Wisconsin to find out the average height of a college student. Of 692 students surveyed, 421 replied that they were over 6 feet tall. What is the standard error? Answer choices are rounded to the hundredths place.
A survey was conducted at local colleges around Madison where 692 students were surveyed, and 421 replied that they were over 6 feet tall showing a standard error of 0.0084 in the average height of a college student.
The standard error is given by the formula given below:
[tex]$$SE= {s}/{\sqrt{n}}$$[/tex]
Where s is the standard deviation,
n is the sample size.
Now let us find out the standard deviation by using the formula given below:
[tex]$$s=\sqrt{\frac{(421-271.17)^2+(271.17-270)^2}{692-1}}$$[/tex]
After calculating we get that the standard deviation s is equal to $0.2208$.
Now let us plug the value of the standard deviation s and sample size n into the formula for standard error:
[tex]$$SE={s}/{\sqrt{n}}$$[/tex]
On substituting the respective values, we get [tex]$$SE={0.2208}/{\sqrt{692}}$$[/tex]
On solving, we get that the standard error is equal to 0.0084
Therefore, the standard error is 0.0084.
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The moment of inertia of a solid cylinder about its axis is given by 1/2MR 2 . If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is:A. 1:1
B. 2:2
C. 1:2
D. 1:3
Answer:
I = 1/2 M R^2 moment of inertia
Translational energy due to rotation
Er = 1/2 I ω^2 = 1/2 M R^2 ω^2 = 1/2 M V^2 since V = R ω
Thus (A) the translational KE is equal to the rotational energy and
Ek = Er + Et for the total energy of the cylinder
what is the acceleration of an object flying upward during free fall?
During free fall, an object is subject to the force of gravity and its acceleration is equal to the acceleration due to gravity (g), which is approximately 9.81 meters per second squared (m/s²) near the surface of the Earth.
If an object is flying upward during free fall, its acceleration will still be equal to -9.81 m/s² (note the negative sign indicating that the acceleration is downward). This is because the direction of the acceleration due to gravity is always toward the center of the Earth.
Even if an object is moving upward, it is still subject to the gravitational force, which causes it to decelerate until it reaches its highest point and then starts to fall back down.
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which of the following actions will cause the relative humidity of an air parcel to increase? select all that apply
a. Keep the parcel’s temperature constant and increase the parcel’s dew point
b. Decrease the parcels temperature and increasethe parcels dew point
c. Keep the parcel’s temperature constant and keep the parcels dew point constant
d. Increase the parcels temperature and increase the parcels dew point
e. Keep the parcels dew point constant and increase the parcels temperature
The relative humidity of an air parcel will increase if any of the following actions are taken:
Keep the parcel’s temperature constant and increase the parcel’s dew pointDecrease the temperature of the parcel and increase the parcels dew pointIncrease the temperature of the parcel and increase the parcels dew pointKeep the parcels dew point constant and increase the temperature of the parcelWhat is relative humidity?To understand this further, we can look at the formula for relative humidity, which is the amount of water vapor in the air divided by the amount of water vapor that can exist at a particular temperature. When the temperature is kept constant and the dew point increases, the amount of water vapor in the air increases, resulting in an increase in relative humidity.
The followings are the given options and the actions they will take that will cause the relative humidity of an air parcel to increase:
Option A: Keep the parcel's temperature constant and increase the parcel's dew point. This action would increase the RH of the air parcel because it will increase the quantity of water vapor in the air parcel. As the parcel's temperature is constant, the ability of the air to hold water vapor also remains constant.
Option B: Decrease the parcel's temperature and increase the parcel's dew point. This action would also increase the RH of the air parcel. As the temperature of the parcel decreases, the amount of moisture that the air can contain also decreases. When the dew point is raised, the quantity of water vapor in the air parcel rises relative to the amount it can carry.
Option C: Keep the parcel's temperature constant and keep the parcel's dew point constant. In this case, there will be no increase in RH because the quantity of water vapor in the air parcel will remain the same as the ability of the air to hold water vapor remains constant.
Option D: Increase the parcel's temperature and increase the parcel's dew point. Increasing the parcel's temperature will raise the ability of the air to hold water vapor, but it will not increase the amount of water vapor in the air parcel. As a result, the RH of the air parcel will decrease.
Option E: Keep the parcel's dew point constant and increase the parcel's temperature. This action will also decrease the RH of the air parcel as it will increase the amount of moisture that the air can hold. Thus, the relative humidity will decrease.
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Let the mass of the sled be m and the magnitude of the net force acting on the sled be Fnet . The sled starts from rest. Consider an interval of time during which the sled covers a distance s and the speed of the sled increases from v1 to v2 . We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled. Use W = F s cos (theta) to find the net work Wnet done on the sled. Express your answer in terms of some or all of the variables m ,v1 and v2 .
Total work done is Wnet = 1/2mv₂² - 1/2mv₁²
Let the mass of the sled be m and the magnitude of the net force acting on the sled be Fnet .
The sled starts from rest. Consider an interval of time during which the sled covers a distance s and the speed of the sled increases from v₁ to v₂ . We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled.
Use W = F s cos (theta) to find the net work Wnet done on the sled. Express your answer in terms of some or all of the variables m ,v₁ and v₂.Using the work-energy principle, we can calculate the work done on an object in terms of its change in kinetic energy. Consider the sled being acted upon by a force Fnet.
W = ΔK is used to calculate the work done on the sled as it moves from rest to velocity v₁ and then to velocity v₂ over a distance s.
Considering the sled to be the system under study, we can write the net work done on the sled as Wnet = ΔK.Wnet = 1/2mv₂² - 1/2mv₁² = Fnet s cos θWnet = Fnet s cos θ = 1/2mv₂² - 1/2mv₁²
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a 3 3-inch candle burns down in 12 hours. if b represents how much of the candle, in inches, has burned away at any time given in hours, t, write a proportional equation for b in terms of t that matches the context.
The proportional equation that matches the context of a 33-inch candle burning down in 12 hours is b = 2.75t.
A candle that is 33 inches long is called a 33-inch candle. Candles are a popular decorative item that is commonly used for lighting, as decoration for weddings, and parties, or to create an aromatic atmosphere. B represents the length of the candle that has burned away at any time given in hours, t.
To find the proportional equation for b in terms of t that matches the context of a 33-inch candle burning down in 12 hours, the following steps should be followed:
Identify the given informationThe length of the candle (l) = 33 inchesThe time taken for the candle to burn down (t) = 12 hours
Determine the rate of burning The rate of burning of the candle is given by l/t. Therefore, the rate of burning = 33/12 = 2.75 inches per hour.
The proportional equation for b in terms of t is given by b = rt where r is the rate of burning. Therefore, b = 2.75t.
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Find the value of x. Round the length to the nearest tenth.
Answer:
well first u divide the numbers
2.1 [2] As more resistors are added in series, the equivalent resistance of the circuit approaches infinity. In contrast, as more resistors are added in parallel, the equivalent resistance a. approaches infinity b. approaches zero c. becomes zero d. approaches 1 Ω
2.2 [2] Kirchhoff's loop rule is equivalent to which of the following principles? a. conservation of charge b. conservation of energy c. conservation of mass d. conservation of force
2.1 As more resistors are added in parallel, the equivalent resistance approaches zero
2.2 Kirchhoff's loop rule is equivalent to the conservation of energy principle.
As more resistors are added in series, the equivalent resistance of the circuit approaches infinity. In contrast, as more resistors are added in parallel, the equivalent resistance approaches zero. This statement is TRUE. The equivalent resistance, Req, of a parallel combination of resistors is less than any of the resistors in the combination, while for a series combination it is equal to the sum of the resistances.
Kirchhoff's loop rule is equivalent to the conservation of energy principle. Kirchhoff's loop rule or Kirchhoff's voltage law (KVL) is a result of the conservation of energy principle. The principle of conservation of energy states that energy can neither be created nor destroyed, it can only be transformed from one form to another. In a closed loop, the total energy gained is equal to the total energy lost, according to the principle of conservation of energy.
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The electric potential at a distance d
from a certain point charge is V relative to infinity. What is the potential (relative to infinity) at half the distance for the same charge?
A. V/4
B. 2 V
C. V/2
D. 4 V
The electric potential from a certain point charge when the distance is halve for the same charge will be V/2. Thus, the correct option will be C.
According to the Coulomb's law, the electric field is the gradient of the electric potential. And, the electric potential V is given by:V = kQ/r, where Q is the charge, r is the distance between the charge and the point where the potential is being calculated, and k is Coulomb's constant. Here, the electric potential at a distance d from a certain point charge is V relative to infinity.
The electric potential (relative to infinity) at half the distance for the same charge is the distance r/2, so:
V' = kQ/r
2V' = kQ/(d/2)
V' = 2kQ/d
V' = V/2
Therefore, the electric potential at half the distance for the same charge is V/2.
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what device is used through ureteroscope to capture an inact calculus or fragments if fractured by laser
The device used through a ureteroscope to capture an intact calculus or fragments if fractured by laser is called a basket retrieval device.
A ureteroscope is a specialized tool that is used to examine and treat the inside of the ureter and kidney. It is made up of a long, thin tube with a camera and a light source at the end, which is inserted into the patient's urinary tract through the urethra. The physician will be able to examine the lining of the bladder, ureters, and kidneys during this examination.
A basket retrieval device is a specialized tool that is used during ureteroscopy, which is a minimally invasive surgical technique used to examine the inside of the urinary tract. It is used to remove kidney stones or any fragments that have been broken down by laser lithotripsy.The basket retrieval device works by capturing the stones or fragments with its metal "basket" and then removing them from the body. The physician will then be able to extract the stones or fragments by retracting the basket into the ureteroscope's working channel. The stones will be disposed of or sent to a lab for further testing.
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if the leftover energy in the previous problem is 134.9 j (it's not, don't go back and try to use this value) and the mass is 2 kg, what speed (in m/s) does the block have at the bottom of its slide? revisit the definition of ke if needed.
The speed of the block at the bottom of its slide is 16.4 m/s.
In the previous problem, the kinetic energy of the block was found to be 135 J.
The formula for kinetic energy is
KE = 1/2mv²,
Where:
m is the mass of the object and v is its velocity.Now we can use the same formula to find the velocity of the block at the bottom of its slide.
KE = 1/2mv²
We know that the mass of the block is 2 kg, and the kinetic energy at the end of the slide is 135 J.
KE = 135 Jm = 2 kg1/2mv² = 135 Jv² = 2(135 J) / 2 kgv² = 270 JV = sqrt(270 J) / 2 kgV = 16.4 m/s
Therefore, the speed of the block at the bottom of its slide is 16.4 m/s.
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a 421 kg block is puled up a 4.54 degree incline by a constant force f of 3282 n. the coefficient of friction mu between the block and the plane is 0.47. how fast in m/s will the block be moving 6 seconds after the pull is applied?
The block will be moving at 3.97 m/s 6 seconds after the pull is applied.
Given Mass of the block, m = 421 kg, Inclined angle, θ = 4.54°, Force applied, F = 3282 N, Coefficient of friction, μ = 0.47, Time, t = 6 s
Using Newton's second law of motion, F - μmg sin θ = ma
Where,
m = Mass of the block
g = Acceleration due to gravity
a = Acceleration of the block
Substituting the given values,
3282 - 0.47 × 421 × 9.81 × sin 4.54° = 421 × a
a = 0.6614 m/s²
Using kinematic equations of motion,
v = u + at
Where,
u = Initial velocity
v = Final velocity
a = Acceleration
t = Time
Since the initial velocity is zero, the above equation becomes
v = at
Substituting the values,
v = 0.6614 m/s² × 6 s
v = 3.97 m/s
Therefore, the block will be moving at 3.97 m/s 6 seconds after the pull is applied.
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