Through an orbit what remains constant ?

a. Kinetic energy
B. Potential energy
C. Total energy
D. Work

Answers

Answer 1

Answer:

c. Total energy

Explanation:

An "orbit" is a path for an object to follow. An example of an object is the "satellite."

At certain points in the orbit, the satellite increases its speed and decreases its speed in relation to the gravity caused by the Earth. As it moves farther away from the Earth, its speed slows down. So, this means that the kinetic energy changes. It also gains and loses height which is responsible for the changes occurring regarding potential energy. This is true for elliptical motion of satellites.

However, the total mechanical energy (TME) of the satellite remains the same or is unchanged (elliptical/circular).


Related Questions

Do you agree or disagree with Kinsey that sexuality is a fluid concept, ranging on a seven-point scale from homosexual to heterosexual? Or, do you think these are discrete constructs that have finite points where they begin and end sexuality?

Answers

Answer:

kinsey sexuailty doesn't matter because that is his or her  

Explanation:

Total internal reflection will occur when:

light goes from high to low density above the critical angle
light goes from low to high density above the critical angle
light goes from high to low density below the critical angle
light goes from low to high density below the critical angle

Answers

Answer:

Try B or C if I'm wrong sorry

Explanation:

Select the correct answer.
Proper technique can help prevent injuries.
A.
True
B.
False

Answers

Answer:

b

Explanation:

Answer:

false

Explanation:

Brain injuries can be caused by strong linear acceleration of the head. Imaging measurements of a human head upon impact show that the magnitude of displacement of the brain relative to the skull is, at most, approximately 1 mm, independent of the magnitude of the acceleration. Studies indicate that the probability of mild traumatic brain injury (MTBI) is approximately 25% for linear accelerations of the head of 50g, where g is the acceleration due to gravity (g = 9.8 m s 2 ). The probability of MTBI is ≈100% for linear accelerations of 150g. For this problem, model the case of a bicycle accident as a human head falling straight down from rest at a height of 2 meters. Assume the ground is very hard and does not compress upon impact. Also assume constant deceleration of the brain upon impact.(a) For the case when the cyclist is not wearing a helmet, what is the deceleration that the cyclist’s head undergoes upon contact with the ground? Express your answer in g’s. What is the probability of suffering a MTBI?(b) Suppose you want to design a helmet that reduces the probability of MTBI to less than 25%. By how much must the helmet compress after impact to achieve this goal for our model assumptions? Express your answer in cm. Do typical bicycle helmets provide this amount of compression?(c) Discuss ways in which our model (a human head falling straight down from rest at a height of approximately two meters onto an incompressible surface) could overor under-estimate the probability of MTBI in a bicycle accident.

Answers

Answer:

a) a = 2000 g ,  b)  y = 4 cm

Explanation:

a) We can solve this part using the free fall relations

           

Let's start by finding the speed with which it reaches the ground, just before it starts to decelerate

           v₁² = v₀² + 2 g y

as it comes out of rest the initial velocity is v₀ = 0

           v₁ = √ 2g y

           v₁ = √ (2 9.8 2)

           v₁ = 6.26 m / s

in the braking part the distance is y₂ = 1 mm = 0.001 m

          v² = v₁² - 2 a y₂

       when stopping its velocity is zero v = 0

        a = v₁² / 2y₂

        a = 6.26² / (2 0.001)

        a = 1.96 10⁴ m / s²

This is the braking acceleration.

Let's look for its relationship with the acceleration of gravity

          a / g = 1.96 10⁴ / 9.8

          a / g = 2000

          a = 2000 g

b) to reduce the injury to less than 25%, the maximum acceleration that must be a = 50g = 490 m / s²

Let's find the distance it must travel to have this acceleration

          v² = v₁² - 2 a y

          y = (0 + v₁²) / 2a

          y = 6.26² / (2 490)

          y = 0.04 m

          y = 4 cm

therefore we must have a helmet that deforms 4 cm before the head stops

Most bicycle helmets provide this protection since they are quite elongated

c) The main error in the calculations is not taking into account the  

resistance with the air, in this sense the speed would be less than the calculated one and our calculations would be over estimated.

The second factor is that in a fall there may be an initial velocity different from zero, so the velocity would be higher and the acceleration as well, in this case our calculations are underestimated.

Sara pushes her 45 kg younger sister with a force of 300 N to the right and encounters 120 N of friction. Calculate the little sister’s acceleration. Show steps please:)

Answers

The Net force is 280 N (300-120). The acceleration = force / mass = 280 / 45 = 6.2 m/s^2

5. How long does it take a car to cross a 30 m wide intersection after the light turns green after waiting at a traffic light, assuming the car accelerates at a constant 2.0 m.s2. Ignore the driver’s reaction time

Answers

Answer:

S = Vo t + 1/2 a t^2 = 1/2 a t^2       V0 = 0 for a standing start

t = (2 S / a)^1.2

t = (2 * 30 / (2.0) =  (30)^1/2 = 5.5 sec

PHYSICS 1403
Lab Homework - Friction on a Ramp
A laborer wants to move crates containing bottles of olive oil from a truck to the ground by sliding them
along a ramp. The ramp is 6 m long and is at an angle of 25º. There is friction on the ramp for the first
crate. The laborer doesn't know that there is a small leak in one of the bottles. The leak leaves a layer of
oil on the ramp. The oil creates a frictionless surface for the second crate Wayne sends down the ramp. At the bottom of the ramp, the speed of the second crate (without friction) is 2.5 the speed of the first crate (with friction). Find the coefficient of kinetic friction. Hint: this is a multistep problem that is
be solved using only energy equations. Do not use kinematics or you will not receive full
credit, even if your answer is correct. Use conservation of energy and start with the frictionless case.

Answers

Hi there!

Hi there!

We can begin by simplifying the work-energy theorem for Crate 2.

Since there is no friction, there is no energy dissipated. Thus, the initial energy is equal to the final energy.

Initially, we only have gravitational potential energy (U = mgh), and when the box has fully slid down, it only has kinetic energy (KE = 1/2mv²), therefore:
[tex]E_i = E_f\\\\mgh = \frac{1}{2}mv^2[/tex]

We can cancel out the mass and solve for velocity.

[tex]gh = \frac{1}{2}v^2\\\\v^2 = 2gh \\\\v = \sqrt{2gh}[/tex]

We must use right triangle trigonometry to solve for the HEIGHT given the ramp's length (hypotenuse).

We can use sine:
[tex]sin\theta = \frac{\text{h}}{L} \\\\Lsin\theta = h = 6 * sin(25) = 2.5357 m[/tex]

Now, solve for velocity.

[tex]v = \sqrt{2(9.8)(2.5357)} = 7.05 \frac{m}{s}[/tex]

Since this is 2.5 times the speed of the first crate, we know that the final velocity of crate 1 is:


[tex]v_1 = \frac{v}{2.5} = 2.82 \frac{m}{s}[/tex]

Crate 1:
In this instance, we have friction. Recall the following.

[tex]F_f = \mu N[/tex]

On an incline, the normal force is equivalent to the cosine of the force of gravity, so:
[tex]N = mgcos\theta[/tex]

Now, create an equation for the force due to friction.

[tex]F_f = \mu mgcos\theta[/tex]

The work done by any force is:
[tex]W = F \cdot d\\\\W_f = \mu mgdcos\theta[/tex]

In this instance, d = the ramp's length, or 6 m.

Now, we can use the work-energy theorem.

Ei = Ef

However, there is energy dissipated; we can call this Wf (Work due to friction). Therefore:
Ei - Wf = Ef

Now, we can rearrange to solve for Wf:
Ei - Ef = Wf

Like above, there is initially only GPE (U = mgh) and finally only KE (K = 1/2mv²), so:
[tex]mgh - \frac{1}{2}mv^2 = \mu mgdcos\theta[/tex]

Solve for the coefficient of friction. Begin by canceling out the mass and multiplying all terms by 2:
[tex]2mgh - mv^2 = 2\mu mgdcos\theta\\\\2gh - v^2 = 2\mu gdcos\theta\\\\\mu = \frac{2gh - v^2}{2gdcos\theta}\\\\\mu = \frac{2(9.8)( 2.5357)- (2.82)^2}{2(9.8)(6)cos(25)}[/tex]

Evaluate:
[tex]\boxed{\mu = 0.39}[/tex]

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionless surface below. Suppose you pull the mass so that the spring is stretched out (initial state) and then you release it, so that the mass starts moving towards the spring is unstretched position (final state). The impulse imparted on the spring-mass system by the force that the wall exerts on the spring is zero, since the wall does not move during this process.

Required:
What total percentage of the period does the mass lie in these regions?

Answers

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

Answers

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60  hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

which vector is the sum of the vectors shown below ?

Answers

Answer:

B

Explanation:

explain how warm air temperatures affect the density of the air​

Answers

Explanation:

As the molecules heat and move faster, they are moving apart. So air, like most other substances, expands when heated and contracts when cooled. Because there is more space between the molecules, the air is less dense than the surrounding matter and the hot air floats upward.

why can't a transformer operate on a DC signal

Answers

Answer:

Transformers do not allow DC input to flow through. This is known as DC isolation. This is because a change in current cannot be generated by DC

Explanation:

Hope this helps

Answer:

a change in current cannot be generated by DC;

7. Explain Why is the
gravitational force that
a friend exerts on you less
than the gravitational force
exerted on you by Earth?

Answers

F=mv^2 / R , mass friend < mass earth so attraction g > friend

Explanation:

F=GMm/d^2

Distance between friends and us <<<<<Distance between centre of earth and us

A student placed a ladder up against a wall as shown below. The normal force applied by the wall in the ladder will be directed:
Up and to the right
To the left
Down and to the left
Straight up

Answers

The normal force is always perpendicular to the surface. So it would be straight to the left of the wall

The normal force applied by the wall in the ladder will be directed to the left. Hence, option (B) is correct.

What is normal force?

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

In this context, normal refers to perpendicular in a geometric sense rather than "usual" or "expected" as it does in everyday parlance. Gravity acts on a person standing stationary on a platform; gravity would otherwise draw the person down into the Earth's core absent a countervailing force from the resistance of the platform's molecules, known as the "normal force."

As normal force acts perpendicular to the surface, the normal force applied by the wall in the ladder will be directed to the left.

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A bar magnet is held vertically with its upper end a little bit below the center of a horizontal metal ring. The upper end of the magnet is its north pole, as shown in the figure. The bar magnet is now dropped. An observer views the ring from above its center. To this observer, how will the induced current in the ring behave as the magnet falls?.

Answers

Answer:

The current will flow counter-clockwise and be decreasing

Explanation:

Please help I only have 20 min left!!!!

Power: The work performed as a function of time for a process is given by W = at3, where a = 2.4 J/s3. What is the instantaneous power output at t = 3.7 s?
Group of answer choices

99 W

139 W

208 W

69 W

Answers

The instantaneous power output at t = 3.7 s right answer is 99 W.

Power is defined by the amount of energy transferred over a period of time. Instantaneous power, on the other hand, refers to the power consumed at a point in time. Instantaneous power is an important metric in electronics. Instantaneous power is the power measured at a specific point in time.

The basic difference between average effort and instantaneous effort is that average effort is the ratio of total work time to total time. Although the instantaneous power is the limit of the average power. The statement of work does not state that the same amount of work he will complete in a second or an hour. Instantaneous power can be positive or negative. Positive instantaneous power means that energy is flowing from the source to the load, and negative instantaneous power means that energy is flowing from the load to the source.

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Two objects have a gravitational force between them of 750N. The two objects now move towards
each other, the gravitational force between them
a. remains at 750N
b. increases above 750N
C. decreases below 750N

Answers

Answer:

b. increases above 750N

Explanation:

Okay

With what force will a car hit a tree if the car has a mass of 3,454 kg and it is accelerating at a rate of 5 m/s2?

Answers

Answer:

F = 17270 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by the acceleration.

ΣF = m*a

where:

F = force [N]

m = mass = 3454 [kg]

a = acceleration = 5[m/s^2]

Now replacing:

F = 3454*5

F = 17270 [N]

The system is released from rest with the spring initially stretched 3 in. Calculate the velocity of the cylinder after it has dropped 0.5 in.

Answers

The velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

Conservation of energy

The velocity of the cylinder can be determined by applying the principle of conservation of energy as shown below;

Ei = Ef

¹/₂k(xi)² + mgh = ¹/₂mv² + ¹/₂k(xt)²

where;

k is spring constant = 6 lb/in = 1050 N/mm is mass of the system = 100 lb = 45.35 kgxi is initial extension = 3 in = 0.076 mxt is the total extension = 2xi + 0.5 in = 6.5 in = 0.165 mh is height = 0.5 in = 0.0127 m

¹/₂k(xi)² = -mgh + ¹/₂mv² + ¹/₂k(xt)²

¹/₂(1050)(0.076)² = -(45.35)(9.8)(0.0127) + ¹/₂(45.35)v² + ¹/₂(1050)(0.165)²

3.032 = -5.64 + 22.68v² + 14.29

3.032 = 8.65 + 22.68v²

-5.62 = 22.68v²

|v| = 0.497 m/s

Thus, the velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

The complete question is below:

The spring has a stiffness of 6 lb/in and the mass of the load is 100 lb.

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A particle is projected vertically upwards with a velocity of 34.3 m/s. For how long is the particle at a height of 49 m and above the point of projection?

Answers

Calculate the time to reach a height of 49 m:

S = V0 t - 1/2 g t^2

S = 34.3 t - 1/2 * 9.8 * t^2

4.9 t^2 - 34.3 t + 49 = 0

I get t = 2 and t = 5    as solutions of the quadratic.

We know that it takes v / a = 34.3 / 9.8 = 3.5 sec to reach maximum height.

So we have:

2 sec to reach 49 m

3.5 sec to reach max height

1.5 sec from 49 m to max height

1.5 sec from max height back down to 49 m

2 sec to fall from 49 m to ground

2 sec + 1.5 sec + 1.5 sec = 5 sec  to go from ground to max height and back down to 49 m

what is a everyday activities examples of newtons 1 law of motion

Answers

Answer:

bouncing  a baskletball

Explanation:

Answer:someone kicking a soccer ball.

Explanation:

Because the ball isn’t in motion until acted on by another object (the foot)

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attached to the end of the unstretched spring and then released. The fish moves downward until the spring is fully stretched, then starts to move back up as the spring begins to contract.
What is the maximum distance through which the fish falls?

Answers

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

The maximum distance through which the fish falls is 0.25m.

What is law of conservation of energy?

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

Given parameters:

Spring constant: k = 200 N/m.

Mass of the fish: m =2.6 kg.

Let the maximum distance through which the fish falls, i.e. the maximum stretched distance of the spring = x.

From law of conservation of energy:

Change in gravity potential energy  = change in potential energy

mgx = 1/2kx^2

2.6×9.8×h = 1/2*200*h^2

h(100h - 2.6×9.8) = 0

h = 0 or h = 2.6×9.8/100 = 0.25m

As h = 0 is the equilibrium position; the maximum distance through which the fish falls is 0.25m.

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Mrs. LaCross leaves school and accidentally leaves her coffee mug on the roof of her car as shown in the picture below.
She was traveling at a constant pace until a student rushes in front of her not using the sidewalks and crosswalks and she had to slam on her brakes. The mug will then move foward

Why will this happen to her coffee mug?

Answers

Answer:

When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake

Explanation:

I'm not entirely sure this is what you were looking for but I hope this helped!

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When a moving object collides with an object that isn’t moving , what happens to the kinetic energy of each object?

Answers

Answer:

The moving object transfers kinetic energy with the object that isn't moving.

Explanation:

Kinetic energy is a form of energy that a object  or particle has by the reason of motion. Potential energy is a form of energy that a object or particle has by the reason of no movement. The object that is moving knocks down or collides with an object that isn't  moving, and the object that is moving transfers kinetic energy to potential energy, and the object that wasn't moving transfers potential energy to kinetic energy.

1. While cruising along a dark sketch of highway at a speed of 25 m/s, you see that a bridge ahead has been washed out. You apply the brakes and uniformly slow down to a stop in 5.0 seconds.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

2. A poorly tuned Yugo can accelerate from rest to a speed of 28 m/s in 20 s.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

3. A bear is sitting at rest at t=0s. At t=5s, the bear notices honey 16 m away and takes off from rest accelerating at a rate of 2 m/s2 for 4 seconds to reach the honey.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

4. A dog runs down his driveway with an initial velocity of -5 m/s for 8 seconds, then uniformly increases his speed to -10 m/s in 5 seconds.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

5. You are driving on the highway at a rate of 40 m/s for 10 seconds when you notice a cop in front of you. Over the next 5 seconds you uniformly slow down to 35 m/s to avoid getting a speeding ticket.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

6. You are traveling 20 m/s when the stoplight in front of you turns red. You step on your break to uniformly slow down to a rest in 5 seconds. You are stopped at the red light for 3 seconds when the light turns green. You speed back up to 20 m/s over the next 5 seconds.

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

The velocity values are ________. (Positive/Negative)

The acceleration value is _______. (Positive/Negative)

Therefore the object must be _________. (Speeding Up/Slowing Down)

Answers

1.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

2.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

3.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

4.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

5.

The velocity values are Positive.

The acceleration value is Positive.

Therefore the object must be Slowing Down.

6.

The velocity values are Negative.

The acceleration value is Negative.

Therefore the object must be Speeding Up.

((I tried my very best I am sorry if any answers I have given are incorrect, but I wish you good luck on your Test/Quiz or whatever you may be currently working on at the time. This took a while.))

\A thin hoop with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 3.5 rad/s. What is its kinetic energy

Answers

The rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

Given the data in the question;

Mass of hoop; [tex]m = 3.0kg[/tex]Radius; [tex]r = 10cm = 0.1m[/tex]Angular speed; [tex]w = 3.5rad/s[/tex]

Rotational or kinetic energy; [tex]E_{rotational} = \ ?[/tex]

Rotational energy or angular kinetic energy

Rotational energy or angular kinetic energy is simply kinetic energy due to the rotation of a rigid body.

It is expressed as;

[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]

Where [tex]I[/tex] is the moment of inertia around the axis of rotation and [tex]w[/tex] is the angular speed or velocity.

For the moment of inertia around the axis of rotation.

[tex]I = \frac{1}{2}mr^2[/tex]

Hence

[tex]E_{rotational} = \frac{1}{2}(\frac{1}{2}mr^2)w^2 \\\\E_{rotational} = (\frac{1}{4}mr^2)w^2[/tex]

Now, we substitute our given values into the above equation to find the rotational or kinetic energy.

[tex]E_{rotational} = (\frac{1}{4}*3.0kg * (0.1m)^2) * (3.5rad/s)^2 \\\\E_{rotational} = 0.0075kgm^2 * 12.25rad/s^2\\\\E_{rotational} = 0.092kg.m^2/s^2\\\\E_{rotational} = 0.092J[/tex]

Therefore, the rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

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g A satellite is orbiting planet Earth with a linear speed of 3,914 m/s. The orbital radius of the satellite in km is:

Answers

The orbital radius of the satellite orbiting planet Earth at the given linear speed is 26,000 km.

Orbital radius of the planet

The orbital radius of the planet is the distance of the planet from the center of the Earth. The orbital radius of the planet is calculated as follows;

[tex]v = \sqrt{\frac{GM}{r} }[/tex]

where;

M is mass of EarthG is gravitational constantr is the orbital radius

r = GM/v²

r = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴)/(3,914²)

r = 2.6 x 10⁷ m

r = 26,000 km

Thus, the orbital radius of the planet is 26,000 km.

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5. A capacitor is discharging through a resistor. The initial voltage across the capacitor is 3.2 V at t = 0. The voltage across the capacitor at time t = 20 ms is 0.8 V. The time it takes for the voltage across the capacitor to drop from 0.8 V to 0.2 V is ​

Answers

Hi there!

Recall the equation for the voltage of a discharging capacitor:

[tex]V_C(t) = V_0e^{-\frac{t}{\tau}}[/tex]

V₀ = Initial voltage of the capacitor (V)
t = Time (s)
τ or RC = Time Constant (s)

With the given information, we can plug in the value for V₀:
[tex]V_C(t) = 3.2e^{-\frac{t}{\tau}}[/tex]

We are given that at t = 20 ms (0.02 s), the voltage of the capacitor is 0.8V. We can use this to solve for the time constant (τ).

[tex]0.8 = 3.2e^{-\frac{0.02}{\tau}}\\\\0.25 = e^{-\frac{0.02}{\tau}}[/tex]

Take the natural log of both sides and solve.

[tex]ln(0.25) = -\frac{0.02}{\tau}\\\\-1.3863 = -\frac{0.02}{\tau}\\\\\tau = \frac{0.02}{1.3863} = 0.0144 s[/tex]

Now, we can use this time constant to solve for the time taken for the voltage to drop from 0.8 V to 0.2 V. Solve for the time taken for the capacitor's voltage to drop to 0.2 V:

[tex]0.2= 3.2e^{-\frac{t}{0.0144}}\\\\0.0625 = e^{-\frac{t}{0.0144}}\\\\ln(0.0625) = -\frac{t}{0.0144}\\\\t = (-2.773)(-0.0144) = 0.04 s[/tex]

Now, subtract the times:
[tex]0.04 - 0.02 = 0.02 = \boxed{20 ms}[/tex]

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? 214 nm 224 nm 204 nm 100 nm 234 nm

Answers

Answer:

The correct option is a: 214 nm.

Explanation:

The wavelength (λ) of the light can be calculated as follows:

[tex] \lambda = \frac{yd}{nD} [/tex]   (1)

Where:

y: is the distance of the 7th bright fringe on the screen from the central fringe = 10.0 cm

d: is the distance between the slits = 3.00x10⁻⁵ m

D: is the distance between the screen and the slits = 2.00 m

n = 7

By entering all the above values into equation (1) we have:

[tex] \lambda = \frac{yd}{nD} = \frac{0.10 m*3.00 \cdot 10^{-5} m}{7*2.00 m} = 214.3 nm [/tex]

Therefore, the correct option is a: 214 nm.

I hope it helps you!

Six insulated containers hold 3,750 g of water at 24°C. A small copper cylinder is placed in each container; the masses and initial temperatures of the cylinders vary as given below. Rank the containers according to the maximum temperature of the water in each container after the cylinder is added, from largest to smallest. You may assume that the cylinder is completely submerged in the water.A. m = 250 g; T = 30°CB. m = 500 g; T = 60°CC. m = 750 g; T = 90°CD. m = 500 g; T = 15°CE. m = 750 g; T = 30°CF. m = 250 g; T = 60°C

Answers

Answer:

the order from highest to lowest is      C> B> F> E> A> D

Explanation:

This is an exercise in calorimetry where the heat given off by the copper cylinder is equal to the heat absorbed by the water  

     Q_c = m_Cu ce_Cu ΔT_Cu  

    Q_a = m_water Ce_water ΔT  

    Q_c = Q_a  

    m_Cu ce_Cu (T_o-T_f) = m_water ce_water (T_f - T_i)

 

we clear the final temperature and substitute the values

With this expression we can know the final temperature of the system, let's substitute the values ​​that are constant throughout the calculation

With this expression we can know the final temperature of the system, let's substitute the values ​​that are constant throughout the calculation

a) m = 0.250kg, To = 30ºC  

      T_{fa}=(387 0.250 30 + 376740) / (15697.5 + 387 0.250)  

      T_{fa}= 379642.5 / 15794.25  

      T_{f} = 24.04ºC  

b) m = 0.500 kg, To = 60ºC  

     T_{fb} = (387 0.500 60 + 376740) / (15697.5 + 387 0.500)

     T_{fb} = 388350/15891  

     T_{fb = 24.44 ° C

c) m = 0.750 kg, To = 90ºC  

    T_{fc}= (387 0.750 90 + 376740) / (15697.5 + 387 0.750)  

    T_{fc}= 402862.5 / 15987.75  

    T_{fc} = 25.20ºC  

d) m = 0.500 kg, To = 15ºC  

    T_{fd} = (387 0.500 15 + 376740) / (15679.5 + 387 0.500)

    T_{fd} = 379642.5 / 15891  

    T_{fd} = 23.89ºC

the order from highest to lowest is      C> B> F> E> A> D

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