The power loss in the bearing is 1.36 W shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long.
What is viscosity?The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.
The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.
To calculate the power loss in the bearing, we can use the formula:
P = F × V
To calculate the frictional force, we can use the formula:
F = μ × A × P
Where μ is the coefficient of friction, A is the area of the bearing, and P is the pressure of the lubricant.
First, we need to calculate the pressure of the lubricant:
P = μ × Viscosity × (V/D)
Where D is the diameter of the shaft.
P = (0.02) × 20 × (120/0.1) = 480 N/[tex]m^2[/tex]
Next, we can calculate the area of the bearing:
A = π × (D/2)^2 × L
Where L is the length of the bearing.
A = π × (0.1/2)^2 × 0.15 = 0.001178 [tex]m^2[/tex]
Now we can calculate the frictional force:
F = μ × A × P = (0.02) × 0.001178 × 480 = 0.0113 N
Finally, we can calculate the power loss:
P = F × V = 0.0113 × 120 = 1.36 W
Therefore, the power loss in the bearing is 1.36 W.
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a 0.1 kg tennis ball moving at 5m/s [W] is hit by a racquet. If the final velocity of the ball is 5 m/s [E], what is the change of momentum of the ball?
The change in momentum of the ball is 1.0 kg m/s [E].
Momentum calculations
The change in momentum of the ball can be calculated using the formula:
Δp = p₂ - p₁
where Δp is the change in momentum, p₂ is the final momentum, and p₁ is the initial momentum.
The momentum of an object is defined as the product of its mass and velocity, so we can calculate the initial and final momenta of the ball as follows:
p₁ = m₁v₁ = (0.1 kg)(5 m/s [W]) = -0.5 kg m/s
p₂ = m₁v₂ = (0.1 kg)(5 m/s [E]) = 0.5 kg m/s
Substituting these values into the formula for Δp, we get:
Δp = p₂ - p₁ = (0.5 kg m/s) - (-0.5 kg m/s) = 1.0 kg m/s [E]
Therefore, the change in momentum of the ball is 1.0 kg m/s [E] using Δp = p₂ - p₁.
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The visit to jupiter, what did the scientists think when they saw the tornado?
Giving 10 points to each that answer this!!
NASA's Juno spacecraft has been orbiting Jupiter since 2016, providing scientists with unprecedented views of the gas giant. One of the most striking features of Jupiter is its Great Red Spot, a massive storm that has been raging for centuries.
What is Great Red Spot?
On Jupiter, a huge storm known as the Big Red Spot has been raging for at least 400 years and possibly longer. It is a zone of persistently high pressure that is present in the planet's southern hemisphere. The storm, which is thought to be around two to three times the size of our globe, is so enormous that it could easily devour the entire planet.
When it began orbiting Jupiter in 2016, NASA's Juno probe has given scientists hitherto unheard-of pictures of the gas giant. The Great Red Spot, a huge storm that has been raging for ages, is one of Jupiter's most stunning characteristics. The polar regions of Jupiter are however home to numerous additional storms, including sizable and intricate cyclones.
The scale and intricacy of the cyclones on Jupiter's poles, especially a centre cyclone on the north pole known as "Jupiter's North Pole Hexagon," first astounded scientists. These storms can endure for decades or even centuries and are more larger and better organised than any cyclones on Earth.
It's not apparent to me which tornado you're referring about in particular. But, researchers looking at Jupiter's The variety and complexity of these storms have probably fascinated storms, and they may be attempting to comprehend the underlying physical mechanisms. Jupiter's storms can be studied to learn more about planetary weather, including our own.
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What is the gravitational potential energy of a 9.00kg object that is 3.00m below the ground
relative to a point 5.00m below the ground?
-265J
265J
176J
-176J
The gravitational potential energy of the object is -176J. Option 4 is the answer.
How The answer was obtainedThe gravitational potential energy (U) of an object of mass (m) at a height (h) relative to a reference point where gravitational potential energy is defined to be zero is given by:
U = mgh
where g is the acceleration due to gravity (9.81 m/s^2).
In this problem, the object is 3.00 m below the ground relative to a point 5.00 m below the ground. So, the height of the object relative to the reference point is:
h = -3.00 m - (-5.00 m) = 2.00 m
Note that we use a negative sign for h because the object is below the reference point.
Plugging in the values, we get:
U = (9.00 kg) x (9.81 m/s^2) x (-2.00 m) = -176.22 J
Rounding off to two significant figures, the gravitational potential energy of the object is -176 J. Therefore, the answer is -176J.
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6.
a) State the quadratic function that would be represented by a freely falling ballast bag that starts from a
resting position on a balloon 1000 feet above the ground.
Answer:
h(t) = -16t^2 + 1000
Explanation:
The quadratic function that represents the motion of a freely falling ballast bag that starts from a resting position on a balloon 1000 feet above the ground is given by the equation h(t) = -16t^2 + 1000, where h(t) is the height of the ballast bag in feet at time t seconds after it is released, and -16t^2 represents the effect of gravity on the ballast bag as it falls.
Answer:
-16t^2 + 1000
Explanation:
Friction can sometimes be useful, and sometimes it can be nuisance identify whether friction is useful or not in this situation :
*The brakes on a bike .
*Inside the hub of the wheel on a bike .
*The moving parts of the chain and gears .
Answer & Explanation:
Friction is useful in the following situations:
The brakes on a bike: Friction is necessary in the brakes of a bike as it allows the brake pads to grip onto the rim or disc, which helps to slow down or stop the bike.
Inside the hub of the wheel on a bike: Friction is necessary to prevent the wheel from spinning too freely, which could lead to loss of control or accidents. The hub's bearings require just the right amount of friction to keep the wheel spinning freely but not too fast.
Friction is a nuisance in the following situation:
The moving parts of the chain and gears: Friction between the moving parts of the chain and gears can cause the bike to lose efficiency and require more effort from the rider. This can cause wear and tear on the bike and make it harder to ride.
15 Point Emergency!!! Absurd answers will be reported.
Robert is riding his skateboard around a skate park. At the top of one feature, he has 475 J of kinetic energy, and 2240 J of gravitational potential energy. Ignoring friction, how much total energy will Robert have when he is at the bottom of the feature.
2715 J
1765 J
2240 J
475 J
Answer:
475 J
Explanation:
At the top of the feature, Robert has 475 J of kinetic energy and 2240 J of gravitational potential energy. According to the law of conservation of energy, the total energy of a system is constant, so the sum of the kinetic energy and gravitational potential energy at the top of the feature is equal to the sum of the kinetic energy and gravitational potential energy at the bottom of the feature.
Since Robert is at the bottom of the feature, he has lost all of his gravitational potential energy. Therefore, the total energy he has at the bottom of the feature is equal to his kinetic energy at the top of the feature:
Total energy = Kinetic energy at top = 475 J
Therefore, the answer is 475 J, which is option D.
Sally is painting her room. She lifts the 10-newton paintbrush at a distance of 10 meters.
What are the joules of work being put out?
Pls helpp and fast tyy
Answer:
To calculate the work being done by Sally, we need to use the formula:
Work = Force x Distance x cos(theta)
where:
Force = 10 N (the force required to lift the paintbrush)
Distance = 10 m (the distance Sally lifts the paintbrush)
theta = 0 degrees (since Sally is lifting the paintbrush straight up, the angle between the force and the direction of motion is 0 degrees)
So, the work done by Sally is:
Work = 10 N x 10 m x cos(0)
Work = 100 joules
Therefore, the joules of work being put out by Sally when she lifts the 10 N paintbrush at a distance of 10 meters is 100 joules.
Explanation:
A gardener pulles a mower with a force 500J, is done in moving the mower along the grassland through a distance of 20m. & Find the angle of application of the foce
The angle of application of the force is theta = arccos(500 J / (F * 20 m))
Force calculation.
We can use the work-energy principle to solve this problem. The work done by the gardener's force is equal to the change in kinetic energy of the mower. Since the mower starts from rest, the initial kinetic energy is zero. Therefore, the work done by the gardener's force is equal to the final kinetic energy of the mower:
Work = Change in kinetic energy
500 J = (1/2)mv^2
where m is the mass of the mower and v is its final velocity.
Since we don't know the mass of the mower or its final velocity, we cannot directly solve for the angle of application of the force. However, we can use the fact that the work done by a force is equal to the product of the magnitude of the force, the displacement, and the cosine of the angle between the force and the displacement:
Work = Fd cos(theta)
where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and the displacement.
Substituting the given values, we have:
500 J = F(20 m) cos(theta)
Solving for the cosine of the angle:
cos(theta) = 500 J / (F * 20 m)
Since we don't know the magnitude of the force, we cannot directly solve for the angle. However, we can rearrange the equation to isolate the angle:
theta = arccos(500 J / (F * 20 m))
Therefore, we need to know the magnitude of the force to find the angle of application.
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help pls need answers now
At wavelengths [tex]$644.44 \mathrm{~nm}$[/tex] and[tex]$483.33 \mathrm{~nm}$[/tex], the stars [tex]$\mathrm{A}$[/tex] and B have maximum intensities. temperature of most intense star is 6,502.242K
What is a wavelengths frequency?Temperature of star [tex]$\mathrm{A}$[/tex] is [tex]$\mathrm{TA}[/tex]=4,500 [tex]\mathrm{~K}$.[/tex]
Temperature of star $B$ is TB $=6,000 [tex]\mathrm{~K}$.[/tex]
Wavelength of most intense star is [tex]$\lambda_0=446[/tex]
We have to find wavelengths corresponding to maximum intensity of stars at temperatures [tex]$T \wedge$[/tex]and TB and wavelength of most intense star.
From Wein's law, we have
[tex]$\lambda_{\mathrm{nm}}=\frac{2.90 \times 106}{\mathrm{Tk}_{\mathrm{k}}}$[/tex]
Where [tex]$\lambda_{\mathrm{nm}}$[/tex] is in nanometers and [tex]$\mathrm{T} k$[/tex] is in Kelvin.
For Star A, equation (1) becomes
[tex]$\lambda_{\mathrm{A}}=\frac{2.90 \times 106}{\mathrm{TA}}$[/tex]
Using values in (2), we get
[tex]\lambda_{\mathrm{A}}[/tex] & =[tex]\frac{2.90 \times 106}{4,500 \mathrm{~K}} \\[/tex]& =644.44 [tex]\mathrm{~nm}[/tex]
For Star B equation (1) becomes
[tex]$$\lambda_{\mathrm{B}}=\frac{2.90 \times 106}{\mathrm{Tr}}$$[/tex]
Let the temperature of the most intense star is[tex]$\mathrm{T} 0$.[/tex] Now, equation (1) for most intense star becomes [tex]$\lambda_0=\frac{2.90 \times 106}{T_0}$[/tex]
From above equation, we have
[tex]$\mathrm{T}_0=\frac{2.90 \times 106}{\lambda_0}$[/tex]
Using values in (3), we get
[tex]$$\begin{aligned}\lambda_{\mathrm{B}} & =\frac{2.90 \times 106}{6,000 \mathrm{~K}} \\& =483.33 \mathrm{~nm}\end{aligned}$$[/tex]
Using vales in (4), we get
[tex]\begin{aligned}\mathrm{T}_0 & =\frac{2.90 \times 10_6}{446 \mathrm{~nm}} \\= & 6,502.242 \mathrm{~K}\end{aligned}[/tex]
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a 0.1 kg tennis ball moving at 5m/s [W] is hit by a racquet. If the final velocity of the ball is 5 m/s [E], what is the change of momentum of the ball?
The change in momentum of the ball is 1.0 kg m/s [E].
How to calculated the change in momentum of the ball ?
The change in momentum of the ball can be calculated using the momentum equation:
Δp = pf - pi
Where
Δp is the change in momentum pf is the final momentumpi is the initial momentumThe momentum of an object can be calculated as the product of its mass and velocity:
p = mv
Where
p is momentumm is massv is velocityUsing the given values, we can calculate the initial momentum of the ball:
pi = mvi = (0.1 kg)(5 m/s [W]) = 0.5 kg m/s [W]
Similarly, we can calculate the final momentum of the ball:
pf = mvf = (0.1 kg)(5 m/s [E]) = 0.5 kg m/s [E]
Substituting these values into the momentum equation, we get:
Δp = pf - pi = 0.5 kg m/s [E] - 0.5 kg m/s [W] = 1.0 kg m/s [E]
Therefore, the change in momentum of the ball is 1.0 kg m/s [E].
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Logan is standing on a dock holding onto a rope swing that is =4.10 m long and suspended from a tree branch above. The rope is taut and makes a 30.0∘ angle with the vertical. Logan swings in a circular arc, passing through the bottom of the arc and then releasing the rope when it makes an angle of =13.1∘ with the perpendicular.
If Logan's mass is 79.0 kg how much work grav does gravity do on him up to the point where he releases the rope?
Answer:
2025 J
Explanation:
The work done by gravity on Logan can be calculated using the formula W = mgh, where W is the work done by gravity, m is the mass of Logan, g is the acceleration due to gravity and h is the change in height.
First we need to calculate the change in height. The initial height of Logan can be calculated using trigonometry. The vertical component of the rope length when it makes a 30 degree angle with the vertical is 4.10m * cos(30) = 3.55m.
Similarly, when Logan releases the rope at an angle of 13.1 degrees with the perpendicular (or 90-13.1=76.9 degrees with the vertical), his height above water level will be 4.10m * cos(76.9) = 0.93m.
So his change in height will be 3.55m - 0.93m = 2.62m.
Now we can calculate how much work gravity does on him: W = mgh = (79kg)(9.8 m/s^2)(2.62m) ≈ 2025 J.
So gravity does about 2025 J of work on Logan up to the point where he releases the rope.
Objects 1 and 2 attract each other with an electrostatic force of 18.0 N. If the distance separating Objecls 1 and 2 is changed to one-half the original value (i.e., halved), then what is the new electrostatic force?
If object 1 and 2 attracts each other with an electrostatic force of 18N and is separated by some distance, Then if that distance will be halved then the electrostatic force will be = 18×4=72N.
What is coulombs' law of electrostatic?It states that" The force between two charges is directly proportional to the product of their magnitude and inversely proportional to the square of the distance between the charges and acts along the line joining the two charges."
Mathematically,
F∝q1×q2
F∝1/[tex]r^{2}[/tex]
i.e
F= k [tex]\frac{q1 *q2}{ r^{2} }[/tex]
Where q1, q2= magnitude of the charge in C.
and r = the distance between them in m.
k= constant of proportionality.
Question given data,
F1=18N
let r=x (say)
then F1=18=k [tex]\frac{q1 *q2}{ x^{2} }[/tex].....................(1)
if r is halved i.e r=x/2
now new force
F2=k[tex]\frac{q1 *q2}{ (x/2)^{2} }[/tex]
F2=4×k[tex]\frac{q1 *q2}{ x^{2} }[/tex]
F2=4×F1....................(∵F1=k [tex]\frac{q1 *q2}{ x^{2} }[/tex])
=4×18..............................(∵F1=18)
F2=72N
Hence The force between two charged particles when the distance between them is halved is 72N.
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Can someone helping me!
For this activity create a kinship chart of a celebrity family or fictional family on television or in other media. Choose a person as “ego” and use the anthropological symbols to identify as many of their relatives as you can. Be sure to choose a family with at least three known generations.
Kinship charts, often known as kinship diagrams, show relationships. Similar to a family tree chart or a pedigree chart, you can use a kinship diagram to display your genealogy.
What do you meant by a kinship chart?
An illustration of relationships in a family, community, or culture is called a kinship diagram. Kinship charts resemble family trees in many ways. Therefore, kinship diagrams are used more broadly to comprehend how most families in a culture function rather than identifying individual names or modeling the diagram after one family.
Kinship diagrams enable cultural anthropologists to swiftly sketch out relationships between people throughout the interview process. Also, it offers a way to represent kinship patterns visually in cultures without the use of names, which can be confusing, and it protects people's privacy.
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Draw a schematic diagram to show the ball's projectile motion after leaving the ramp with a horizontal velocity. Label the height it dropped and the horizontal distance.
When a ball leaves a ramp with a horizontal velocity, it will follow a curved path known as a projectile motion. The ball will move both horizontally and vertically, with the horizontal velocity remaining constant and the vertical velocity changing due to the force of gravity.
At the point where the ball leaves the ramp, it will have a certain height and horizontal distance from the starting point. As it moves through the air, it will travel a certain horizontal distance before landing on the ground at a lower height.
What is projectile motion?Projectile motion is a type of motion in which an object is launched or thrown into the air, and then moves under the influence of gravity and air resistance. In projectile motion, the object follows a curved path called a trajectory, which consists of both horizontal and vertical components.
The motion of a projectile can be described in terms of its velocity, acceleration, and displacement. At the instant of launch, the object has an initial velocity that can be broken down into horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
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Water at 100 is taken off the stove and allowed to cool for 10 minutes. In this 10 minutes, the temperature decreased to 65. Given that the room temperature is 21, find the temperature of the water in degrees Celsius after an additional 5 minutes waiting time and on condition that the Newton’s Law of cooling is not violated.
Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.
The rate of cooling of the water can be described as:dQ/dt = -k(T - Troom)
where Q is the amount of heat in the water, t is time, k is a constant, T is the temperature of the water, and Troom is the room temperature.
Using the given information, we can set up the following system of equations:
100 = k(100 - 21) (at t = 0) 65 = k(100 - 21) e⁽⁻¹⁰⁾k (at t = 10)
Solving for k in the first equation gives k = 1/79. Plugging this value into the second equation and solving for T gives T = 36.1 degrees Celsius.
Now, to find the temperature after an additional 5 minutes, we can use the same equation with a new time value:
T = (65 - Troom) e^(-k(10+5)) + Troom
Plugging in the values we know gives T = 30.4 degrees Celsius.
Therefore, the temperature of the water after an additional 5 minutes of cooling is approximately 30.4 degrees Celsius
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URGENT!!! BRAINLIEST WILL BE PICKED!!!
A motor is exerting a clockwise torque on a wheel, with a force of 24.4 N, and a lever arm of 0.22 m. An outside force is exerting a counter-clockwise torque, with a force of 7.9 N, and a lever arm of 0.74 m. Assuming that all forces are perpendicular to their respective lever arms, what will be the resulting motion of the wheel?
There is not enough information to know how the wheel will rotate.
The wheel will not rotate.
The wheel will rotate counter-clockwise.
The wheel will rotate clockwise.
Answer:
Since the net torque is negative, this means that the counterclockwise torque is greater than the clockwise torque. As a result, the wheel will rotate in the counterclockwise direction. Therefore, the answer is:
The wheel will rotate counterclockwise.
Explanation:
To determine the resulting motion of the wheel, we need to calculate the net torque acting on the wheel. Clockwise torque is positive, while counterclockwise torque is negative. Therefore, the net torque can be calculated as follows:
Net torque = (Clockwise torque) - (Counterclockwise torque)
Net torque = (24.4 N) x (0.22 m) - (7.9 N) x (0.74 m)
Net torque = 5.368 Nm - 5.846 Nm
Net torque = -0.478 Nm
two capacitors of capacitance C and 2C are connect in parallel and these are charged up to V volt. If the bartwry is removed and dielectric medium of comstant K is put between the plates of the first capacitor , then the potential on each capacitor is ?
tysm!
Initially, the two capacitors are charged up to the same voltage V, so the charge on the capacitor of capacitance C is Q = CV, and the charge on the capacitor of capacitance 2C is Q' = 2CV.
When the battery is removed and a dielectric medium of constant K is inserted between the plates of the first capacitor (of capacitance C), the capacitance of this capacitor becomes KC. The capacitance of the second capacitor remains the same at 2C.
The total charge on the capacitors remains the same, as no charge is lost or gained during this process. Therefore, we have:
Q = Q' = 2CV
After the insertion of the dielectric medium, the potential difference across the two capacitors will be different, as their capacitances have changed. Let V1 be the new potential difference across the capacitor of capacitance KC, and V2 be the potential difference across the capacitor of capacitance 2C.
We know that the total charge on the capacitors remains the same, so we can write:
Q = C1V1 + C2V2
where C1 = KC is the capacitance of the first capacitor (after insertion of the dielectric), and C2 = 2C is the capacitance of the second capacitor.
Substituting the values of C1, C2 and Q, we get:
CV = KV1 + 2CV2
Simplifying, we get:
V1 = (C/K) V - 2V2
But we also know that the potential difference across the two capacitors must be the same, i.e., V = V1 + V2. Substituting the value of V1 in this equation, we get:
V = [(C/K) V - 2V2] + V2
Simplifying, we get:
V2 = (1/3) V/K
Therefore, the potential difference across the capacitor of capacitance KC (after insertion of the dielectric) is:
V1 = (C/K) V - 2V2 = (5/3) V - (2/3) V/K
Thus, the potential on each capacitor after the insertion of the dielectric is V1 = (5/3) V - (2/3) V/K for the capacitor of capacitance KC, and V2 = (1/3) V/K for the capacitor of capacitance 2C.
Explain how it is possible to hear sound waves, when there is no direct path to the source of the sound.
Sound waves can travel through a medium (such as air, water, or solid objects) and can reflect or diffract around obstacles, allowing them to be heard even when there is no direct path to the source of the sound.
When sound waves encounter an obstacle, they can either reflect or diffract. Reflection occurs when sound waves bounce off a surface and return in the direction they came from. Diffracton occurs when sound waves bend around an obstacle and continue in a different direction. This allows sound waves to travel around corners and through small openings.
For example, if someone is talking behind a wall, the sound waves they produce can diffract around the wall and reach your ears, allowing you to hear them. Similarly, when you hear an echo, it is because sound waves have reflected off a surface and reached your ears.
In some cases, sound waves can even travel through solid objects. This is known as vibration or bone conduction. For example, when you hear your own voice or a piece of music through a set of headphones, the sound waves are transmitted through the headphones and vibrate the bones in your ear, which sends signals to your brain that you perceive as sound.
In summary, sound waves can travel through a medium and reflect or diffract around obstacles, allowing them to be heard even when there is no direct path to the source of the sound.
Need help right now 20 points!!
3. You use a force of 190 N to walk your bike of mass 22 kg over a distance of 2.0 m. The bike
was initially not moving. How fast is it going while you are pushing? {Fd=AKE}
(Initial velocity = 0) by
The bike is going at a speed of 7.76 m/s (or about 28 km/h) while you are pushing it with a force of 190 N over a distance of 2.0 m.
What is work-energy principle ?We can use the work-energy principle to determine the final velocity of the bike. The work-energy principle states that the net work done on an object is equal to its change in kinetic energy. In this case, the net work is done by the force that you apply to the bike over a distance of 2.0 m:
Net work = force x distance x cos(theta) = 190 N x 2.0 m x cos(0) = 380 J
The change in kinetic energy of the bike is given by:
[tex]Delta K = 1/2 mv^2 - 0[/tex]
where m is the mass of the bike and v is its final velocity.
Setting the net work equal to the change in kinetic energy, we have:
[tex]380 J = 1/2 (22 kg) v^2[/tex]
Solving for v, we get:
[tex]v = sqrt(2 x 380 J / 22 kg) = 7.76 m/s[/tex]
Therefore, the bike is going at a speed of 7.76 m/s (or about 28 km/h) while you are pushing it with a force of 190 N over a distance of 2.0 m.
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Exam Instructions
Question 8 of 20:
Select the best answer for the question
8. Which of the following is a healthy example of practicing self-care?
OA. Using junk food to cope with stress
OB. Spending extra hours working at the office
OC. Reflecting on a happy memory
OD. Avoiding friends and family for weeks
Mark for review (Will be highlighted on the review page)
o Previous Question
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Self-care can be demonstrated in a healthy way by recalling a happy memory.
Is self-care included in reflection?The first step toward self-care is allowing yourself to be honest and vulnerable. Through self-reflection, you can at last let out all that you've been keeping down. Your journal is your very own private platform on which you can record your successes, anxieties, and worries.
What good examples of self-care do you have?Taking care of oneself exercises can go from proactive tasks, for example, practicing and practicing good eating habits, to mental exercises like perusing a book or rehearsing care, to otherworldly or social exercises, for example, imploring or getting lunch with a companion.
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Two charges, Q1
= 10 nC
and Q2
= -3.0 nC
, are 15 cm
apart. Find the strength of the electric field halfway between the two charges.
Express your answer with the appropriate units.
Explanation:
E1 = k Q1 / (0.075 m)^2
= 9 x 10^9 Nm^2/C^2 * 10 x 10^-9 C / 0.075^2 m^2
= 400 N/C
E2 = k Q2 / (0.075 m)^2
= 9 x 10^9 Nm^2/C^2 * (-3 x 10^-9 C) / 0.075^2 m^2
= -120 N/C
Since the electric fields due to the two charges are in opposite directions, we can find the net electric field at the midway point by taking the difference between the two:
E = E1 - E2
= 400 N/C - (-120 N/C)
= 520 N/C
7. A man running a 100 m race accelerates uniformly from rest for the first 7 seconds and reaches a velocity of 10 m s. He maintains this velocity for the rest of the race. His time for the race is 12 s. Sketch a velocity-time graph. Calculate (a) the value of T, (b) the acceleration,
(a) Time taken, T is 11.83 s
(b) Acceleration is 1.428 m/s²
What is uniform acceleration ?A body is said to be in uniform acceleration when it is moving in a straight line with an increase in velocity in equal intervals of time.
Here,
s = 100 m
v = u + at
a = (v - u)/t = (10 - 0)/7
a = 10/7 m/s²
(a) s = ut + 1/2 at²
100 = 0 + 1/2 x 10/7 x t²
t² = 140
t = √140 = 11.83 s
(b) Acceleration, a = 10/7 =1.428 m/s²
Hence,
(a) Time taken, T is 11.83 s
(b) Acceleration is 1.428 m/s²
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PLEASE HELP EXPERTS
One type of Personal Noise is:
A. Messages
B. Semantic noise
C. Feedback
D. Prejudice
E. Physical noise
A positive point charge Q1= 2.6×10−5 C is fixed at the origin of coordinates, and a negative point charge Q2= −5.1×10−6 C is fixed to the x axis at x=+2.0m.
Find the location of the place along the x axis where the electric field due to these two charges is zero. (the location should be (x) m from Q2
The location at the point where electric field becomes zero is at 3.6m, when a positive point charge Q1= 2.6×10−5 C is fixed at the origin and a negative point charge Q2= −5.1×10−6 C is fixed at 2.0m.
What is electric field ?Electric field is field around electrically charged particle where columbic force of attraction or repulsion can be experienced by other charged particles. It is denoted by letter E and it's SI unit is V/m Volt per meter or N/C newton per coulomb.
Electric field comes inward to the center of the negative charge and it is going outward for positive charge.
Given,
Q₁= 2.6×10⁻⁵ C
Q₂= −5.1×10⁻⁶ C
x = 2.0m,
In this case there are two charges, one at the origin and other is at 2m distance from origin. Both these charges are kept along positive x axis. looking at the situation there are 2 cases
Case 1 ( electric field in between two charges)
in this case direction of the resultant electric field in between two charges is along positive x direction (direction of electric field of positive charge is outward along x axis which is along positive x direction and direction of electric field of negative charge is inward along x axis which is also in positive x direction).
Case 2 ( right side of the negative charge)
In this case direction of electric field of positive charge is outward along x axis which is along positive x direction and direction of electric field of negative charge is inward along x axis but which is in opposite direction of electric field due to positive charge.
Hence our electric field is zero at right side of the negative charge.
on right side, electric fields due to both charges are opposite to each other then we can write mathematically E₁ - E₂ = 0 at the point where resultant electric field is zero. Where E₁ is elctric field due to positive charge and E₂ is electric field due to negative charge.
E₁ - E₂ = 0
E₁ = E₂
[tex]\frac{kQ_{1}}{x^{2}} = \frac{kQ_{2} }{(x+2)^{2}}[/tex]
[tex]\frac{Q_{1}}{x^{2}} = \frac{Q_{2} }{(x+2)^{2}}[/tex]
[tex]\frac{Q_{1}}{Q_{2} }} = \frac{x^{2} }{(x+2)^{2}}[/tex]
(2.6×10⁻⁵ C ÷ -5.1×10⁻⁶ C) [tex]= \frac{x^{2} }{(x+2)^{2}}[/tex]
[tex]\frac{x^{2} }{(x+2)^{2}}[/tex] = 5.098 ≅ -5.1
x² = 5.1 (x+2)²
x = 2.25(x+2)
x=2.25x + 4.5
x(1-2.25) = 4.5
x(-1.25) = 4.5
x = 3.6m
Hence Electric field is zero at 3.6m from origin.
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calculate the distance covered by you if you travel at the speed of 5 meter per second for 3 minutes
Answer: 900 m
Explanation: If you walk at a speed of 5 meters per second for 3 minutes, or 180 seconds, then you will have walked 900 meters. This can be seen in the equation [tex]x=x_{0}+v_{x0}t+\frac{1}{2} a_{x}t^2[/tex]. As the velocity is constant (no acceleration) and the initial position is zero, plugging in known values for v and t, you can solve for x of 900 m.
Two points in the xy plane have Cartesian coordinates (5.00, −5.50) m and (−7.00, 7.00) m.
Determine the distance between these points.
Two points in the xy plane have Cartesian coordinates (5.00, −5.50) m and (−7.00, 7.00) m.So, the distance between two points in the xy-plane is 17.32 m.
What is Distance?
Distance is a measurement in numbers of the area between two locations or objects. It is a scalar quantity that symbolises the distance between two places in space or the length of a path taken by an object. Depending on the situation and the scale of the items being measured, distance can be expressed in a variety of ways, including metres, kilometres, miles, feet, and so forth.
Displacement, a vector number that denotes the shift in an object's location with respect to a reference point, is frequently distinguished from distance in physics. Displacement, as opposed to distance, which only analyses the size of the path travelled, considers both the distance travelled by an object and its direction of motion.
To find the distance between two points in the xy-plane, we use the distance formula:
d = [tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2}[/tex]
where (x1, y1) and (x2, y2) are the coordinates of the two points.
Using the given coordinates, we have:
x1 = 5.00 m, y1 = -5.50 m
x2 = -7.00 m, y2 = 7.00 m
Substituting these values into the formula, we get:
d = [tex]\sqrt{(-7.00 - 5.00)^2 + (7.00 - (-5.50))^2}[/tex]
d= [tex]\sqrt{(-12.00)^2 + (12.50)^2}[/tex]
d= [tex]\sqrt{(144.00 + 156.25)}[/tex]
d= [tex]\sqrt{(300.25)}[/tex]
d=17.32 m
Therefore, the distance between the two points is approximately:
d = 17.32 m (rounded to two decimal places)
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A certain planet is impacted by a large asteroid, which sends molten rock flying into space. Two spheres of the molten rock are moving in the same direction. One sphere has a mass of 135 kg and a velocity of 95.0 m/s in the positive y-direction. The other sphere has a mass of 290 kg, and a velocity of 24.0 m/s in the positive y-direction. If the two spheres collide, and fuse together, what would be the velocity of the combined sphere?
77.4 m/s in the negative y-direction
59.5 m/s in the positive y-direction
62.2 m/s in the positive y-direction
46.6 m/s in the positive y-direction
Answer:
46.6 m/s in the positive y-direction.
Explanation:
If a clock expends 2 W of power from a 1.5 V battery, what amount of current is supplying
the clock?
If a clock expends 2 W of power from a 1.5 V battery, 1.3A of current is supplying the clock
What is meant byelectric current ?
When an external voltage is given to one of the conductor's ends, the movement of particles known as an electric current begins. As a result, the negatively charged electrons that are drawn to the positive terminal of the exterior voltage are subject to an electric field.
The current through a conductor is proportional to the voltage across the conductor, according to Ohm's rule. As long as the temperature (and other physical factors) stay constant, this is true for many materials, including metals. The connection between voltage, current, and resistance in an electrical circuit can be calculated using Ohm's Law.
P ⇒ IV
I ⇒ P/V ⇒ 2/1.5 i.e. 1.3A
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The large block shown is =16.0 cm
wide, =19.0 cm
long, and =25.00 cm
high with a mass of 4.55 kg.
This block is passing through air (density of air air=1.43kg/m3).
Calculate the terminal velocity of the block if it is traveling downward with a drag coefficient of Γ=0.822
. See the hint panel for the drag force equation.
The terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.
How to calculate terminal velocity?
To calculate the terminal velocity of the block, we need to use the equation for terminal velocity which is:
v_terminal = sqrt((2mg)/(ρAC_d))
where:
m is the mass of the blockg is the acceleration due to gravity (9.81 m/s^2)ρ is the density of the airA is the cross-sectional area of the blockC_d is the drag coefficientFirst, let's convert the dimensions of the block from centimeters to meters:
width = 16.0 cm = 0.16 m
length = 19.0 cm = 0.19 m
height = 25.0 cm = 0.25 m
Next, let's calculate the cross-sectional area of the block:
A = width * length = 0.16 m * 0.19 m = 0.0304 m^2
Now we can substitute the values into the equation for terminal velocity:
v_terminal = sqrt((24.55 kg9.81 m/s^2)/(1.43 kg/m^30.0304 m^20.822))
v_terminal = 2.08 m/s
Therefore, the terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.
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Through how many radians does the minute hand of a clock rotate in 21 min ?
Answer:
1 Minutes =2Pi/60
21 Minutes=2Pi/60*21
so the answer is, 2.198
Explanation:
we know that
there is 60 min one hour
so the there is 2Pi radians
so 21 minutes is 21/60
and we know the clock hand will travel anticlock wise
and a full circle measures 2Pi radians
and we know that one minute =2Pi /60 radians
so, 21 minutes = (21Pi/60)*60
(Pi=3.14)
21 minutes = 2.19