Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.

Answers

Answer 1

Complete Question

The question image is in the first uploaded image

Answer:

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Explanation:

From the question we are told that

Distance b/w Q mid point and P is given as x

Generally the equation for magnitude of the electric field at the point P is given as

[tex]E=\frac{kQ}{d^2}[/tex]

where

[tex]k=\frac{1}{4\pi e_0}[/tex]

[tex]d=x^2-a^2[/tex]

Therefore

[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]

[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]

Therefore equation for magnitude of the electric field at the point P is

[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]

Determine The Magnitude Of The Electric Field At The Point P. Express Your Answer In Terms Of Q, X, A,

Related Questions

18.) Why is it that Prokaryotic cells can divide a lot quicker than Eukaryotic cells?

Answers

Answer:

Binary Fission. The cell division process of prokaryotes, called binary fission, is a less complicated and much quicker process than cell division in eukaryotes. Because of the speed of bacterial cell division, populations of bacteria can grow very rapidly.

HELP PLEASE!!!
Running at 3.0 m/s, Burce, the 50.0 kg quarterback, collides with Max, the 100.0 kg tackle, who is traveling at 6.0 m/s in the other direction. Upon collision, Max continues to travel forward at 2.0 m/s.If the collision between the players lasted for 0.04 s, determine the impact force on either during the collision

Answers

Answer:

10kN

Explanation:

Given data

m1= 50kg

u1= 3m/s

m2= 100kg

u2= 6m/s

v1= 2m/s

time= 0.04s

let us find the final velocity of Bruce v1

from the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

50*3+100*6= 50*v1+100*2

150+600=50v1+200

750-200=50v1

550= 50v1

divide both sides by 50

v1= 550/50

v1=11 m/s

From

F= mΔv/t

for Bruce

F=50*(11-3)/0.04

F=50*8/0.04

F=400/0.04

F=10000

F=10kN

for Max

F=100*(6-2)/0.04

F=100*4/0.04

F=400/0.04

F=10000

F=10kN

Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals

Answers

Answer:I think it’s self monitoring sorry if wrong

Explanation:

Answer:

It self monitoring

Explanation:

I took the test

In the Bohr model of the atom, the electrons are considered to be particles moving in specific and discrete orbits.

a. True
b. False

Answers

Answer:

True

Explanation:

Bohr proposed an atomic model in which;

the electrons found in an atom can only occupy orbits for which the angular momentum is quantized, which results in discrete values of energy level. the electrons in such stationary state or ground state emits no light, but if an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states.  

Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is different and your example

Please help due today!!

Answers

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. The swimmer's... average speed is 0 m/s and average velocity is 0 m/s. average speed is 0.5 m/s and average velocity is 0.5 m/s. average speed is 1 m/s and average velocity is 0 m/s. average speed is 0 m/s and average velocity is 1 m/s.What is the swimmers average speed and average velocity?

Answers

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

Learn more about speed: https://brainly.com/question/7359669

is 0.8 kilograms bigger then 80 grams

Answers

Answer:

Yes

Explanation:

0.8 kilograms is equal to 800 grams

Answer:

Yes, 0.8 kilograms is greater than 80 grams

Explanation:

0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.

Sorry if I'm wrong, correct me.

If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no longer any force on the ball due to the woman's hand), what is the relationship between aaa and ggg, the magnitude of the acceleration of gravity

Answers

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge 1.02 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force?

Answers

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k [tex]k \frac{ q_{1} \ q_{2} }{ r^{2} }[/tex]

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = [tex]F_{bc x}[/tex]

Y axis  

       [tex]F_{y}[/tex]Fy = [tex]F_{ab} - F_{bc y}[/tex]

let's find the magnitude of each force

     [tex]F_{ab}[/tex] = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      [tex]F_{bc}[/tex] = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + [tex]F_{y}[/tex] j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = [tex]\sqrt{ F_{x}^2 + F_{y}^2 }[/tex]

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= [tex]\frac{ F_{y} }{ F_{x} }[/tex]

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.

What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?

Answers

Answer:

V= -3.6*10⁻¹¹ V

Explanation:

Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:

      [tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]

At r= 0.3 m the spherical surface can be written as follows:

       [tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]

Replacing (2) in (1) and solving for E, we have:

      [tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]

Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:

       [tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]

V = -3.6*10¹¹ Volts.

The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]

We can arrive at this answer as follows:

To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:

[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]

Solving these equations will have:

[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]

As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:

[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]

More information about Gauss' law at the link:

https://brainly.com/question/14705081

A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.7 s the rocket is at a height of 93 m.
What are the magnitude and direction of the rocket's acceleration?
What is its speed at this elevation?

Answers

Answer:

The magnitude and direction of the rocket acceleration is 68.89 m/s² upward.

The speed of the rocket at the given elevation is 186 m/s.

Explanation:

Given;

time to reach the given height, t = 2.7 s

height reached, h = 93 m

initial velocity of the rocket, u = 0

The magnitude and direction of the rocket acceleration is calculated as;

h = ut + ¹/₂at²

h = 0 + ¹/₂at²

h = ¹/₂at²

a = 2h / t²

a = (2 x 93) / 2.7

a = 68.89 m/s²

the direction of the acceleration is upward.

The speed at this elevation, V = u + at

V = at

V = 68.89 x 2.7

V = 186 m/s

How does speed and mass effect kinetic energy ?

Answers

Answer:

in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.

Explanation:

Answer: The more mass of an object has, the more Kinetic energy it has.

Explanation:

Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.

Work put into a machine can never be reclaimed.

True
False

Please answer?!

Answers

I am not sure but I think it is false

A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N

of friction. What distance does the box slide?

Answers

Answer:

the distance moved by the box is 70.03 m.

Explanation:

Given;

mass of the box, m = 35 kg

initial velocity of the box, u = 10 m/s

frictional force, F = 25 N

Apply Newton's second law of motion to determine the deceleration of the box;

-F = ma

a = -F / m

a = (-25 ) / 35

a = -0.714 m/s²

The distance moved by the box is calculated as follows;

v² = u² + 2ad

where;

v is the final velocity of the box when it comes to rest = 0

0 = 10² + (2 x - 0.714)d

0 = 100 - 1.428d

1.428d = 100

d = 100 / 1.428

d = 70.03 m

Therefore, the distance moved by the box is 70.03 m.

A 9800 N car traveling at 22 m/s strikes a concrete bridge support and comes to a complete halt in 0.5 sec. (This only counts as one problem.) Determine the magnitude of the force acting on the car. Suppose a barrier, which contained material that is gradually crushed during impact so that the stopping time was increased to 3 s, had surrounded the bridge support. What would be the magnitude of the new force

Answers

Answer:

43955.12N and 7325.85N

Explanation:

Step one:

Given data

Weight of car= 9800N

therefore the mass of the car is

w=mg

m= w/g

m= 9800/9.81

m= 998.98 kg

Speed= 22m/s

duration of impact= 0.5 seconds

we know that

impulse, Ft= mv

F= mv/t

F= 998.98*22/0.5

F=21977.56/0.5

F=43955.12N

Hence the force is 43955.12N

if the time is 3 seconds

then the force will be

F= mv/t

F= 998.98*22/3

F=21977.56/3

F=7325.85N

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m

Answers

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

[tex]H_m=1.65m[/tex]

[tex]H_E=1.16307m[/tex]

Explanation:

From the question we are told that

Mass of ball [tex]M=2kg[/tex]

Length of string [tex]L= 2m[/tex]

Wind force [tex]F=13.2N[/tex]

Generally the equation for [tex]\angle \theta[/tex] is mathematically given as

[tex]tan\theta=\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{F}{mg}[/tex]

[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]

[tex]\theta=73.14\textdegree[/tex]

Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]

Generally the equation for max Height [tex]H_m[/tex] is mathematically given as

[tex]H_m=L(1-cos146.28)[/tex]

[tex]H_m=0.9(1+0.8318)[/tex]

[tex]H_m=1.65m[/tex]

Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as

[tex]H_E=L(1-cos73.14)[/tex]

[tex]H_E=0.9(1+0.2923)[/tex]

[tex]H_E=1.16307m[/tex]

An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12 cm from it, there is a converging lens of the same focal length.
A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?

Answers

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         [tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]

     

we calculate

          [tex]\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}[/tex]

          [tex]\frac{1}{q}[/tex] = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          [tex]m = \frac{h'}{h} = - \frac{q}{p}[/tex]

       

we substitute

          m = [tex]- \frac{-8.488}{29}[/tex]

          m = 0.29

the positive sign indicates that the image is right

In the Faraday pail experiment a metal ball is lowered into a brass pail. In a variation of this experiment, suppose we charge the metal ball with Q and the brass pail with charge 2Q. The ball is slowly lowered into the pail. At no time does the sphere touch the pail. While the ball is inside, the charge on the outside of the pail is _____. The outer part of the pail is then touched/grounded. Then the ball is removed. After the ball is removed, the charge on the outer surface of the pail is ___. Briefly support your answers with reasoning from Gauss's Law.

Answers

Answer:

a)

*The charges of which and the cube are of the same sign..

          Q_exterior = 3 Q

* the charge of the sphere has a different sign than the charge of the cube,

         Q_exterior =  Q

b)   Q = 0

Explanation:

To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.

Let's analyze each situation

a) Suppose that the charges of which and the cube are of the same sign.

When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of

               Q_exterior = 3 Q

   Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,

     Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall

              Q_exterior = + Q

b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge

             Q = 0

As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy

Answers

Answer:

isnt heat transfer

Explanation:

sorry if im wrong

At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himself. Mason pulled the bat with

force of 15 newtons (N) while Alfredo pulled with a force of 20 newtons (N). Why did Alfredo end up with the bat?

A because the force was 5 N in Mason's direction

B. O because the net force was 5 N in Alfredo's direction

c. O because the net force was 15 N in Mason's direction

D.O because the net force was 20 N in Alfredo's direction

Answers

Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.

Answers

Answer:

A. 72000 C

B. 1100 W

C. 26.4 cents.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 10 A

Voltage (V) = 110 V

Time (t) = 2 h

A. Determination of the charge.

We'll begin by converting 2 h to seconds. This can be obtained as follow:

1 h = 3600 s

Therefore,

2 h = 2 h × 3600 s / 1 h

2 h = 7200 s

Thus, 2 h is equivalent to 7200 s.

Finally, we shall determine the charge. This can be obtained as follow:

Current (I) = 10 A

Time (t) = 7200 s

Charge (Q) =?

Q = It

Q = 10 × 7200

Q = 72000 C

B. Determination of the power.

Current (I) = 10 A

Voltage (V) = 110 V

Power (P) =?

P = IV

P = 10 × 110

P = 1100 W

C. Determination of the cost of operation.

We'll begin by converting 1100 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

1100 W = 1100 W × 1 KW / 1000 W

1100 W = 1.1 KW

Thus, 1100 W is equivalent to 1.1 KW

Next, we shall determine the energy consumption of the range. This can be obtained as follow:

Power (P) = 1.1 KW

Time (t) = 2 h

Energy (E) =?

E = Pt

E = 1.1 × 2

E = 2.2 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost 12 cents.

Therefore, 2.2 KWh will cost = 2.2 × 12

= 26.4 cents.

Thus, the cost of operating the range for 2 h is 26.4 cents.

Which statement about technology is true?
Technology improves a scientist’s ability to make observations.
Technology does not have any effect on a scientist’s work.
Technology is necessary for scientists to perform experiments.
Technology only helps scientists analyze data and write reports.

Answers

Answer:

Technology improves a scientist’s ability to make observation

hope this helps

:D

Answer:

Technology improves a scientist’s ability to make observations.

Explanation:

In the measurement 365 cL the “c” stands for the___ And the “ L”stands for the___

Answers
A.base unit and prefix
B.suffix and prefix
C.prefix and suffix
D.prefix and base unit

Answers

I guess it à because on they make more make it sense

Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B with silk. Nylon is used to rub and charge a third piece of amber. Piece A and B are both repelled by the third piece of amber. This means:____.

Answers

Answer:

ieces A and B must also have the same type of charges

Explanation:

In electrostatics, charges of the same sign repel and charges of different signs attract.

If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.

Also part A and C repel each other, therefore they have the same type of charge.

If we use the transitive property of mathematics, pieces A and B must also have the same type of charges

Einstein's equivalence principle says that __________. Einstein's equivalence principle says that __________. everyone measures the speed of light to be equivalent someone traveling at 0.9c will age at the same rate as someone at 0.99c all people see themselves at an equivalent distance to the center of the universe the effects of gravity are exactly equivalent to the effects of acceleration

Answers

Answer:

Einstein's equivalence principle says that __________.

the effects of gravity are exactly equivalent to the effects of acceleration

Explanation:

The equivalence principle is one of the fundamental laws of physics, as enunciated by Einstein.  It categorically states that the gravitational and inertial forces are of a similar nature.  In physics, a gravitational acceleration is the acceleration of an object in a free fall within a space.  The importance of Einstein's Equivalence Principle is explained by his theory of general relativity.  This theory states that mass is the same, whether inertial or gravitational.

According to the Einstein's equivalence principle, the effects of gravity are exactly equivalent to the effects of acceleration.

Einstein's equivalence principle says that  the effects of gravity are exactly equivalent to the effects of acceleration.

What is Einstein's equivalence principle?

Einstein's equivalence principle states that the the force due to gravity and the force of inertia are similar in the nature and there is no need to distinct them.

The inertia force is opposite in direction to accelerating force of a body.

Thus the Einstein's equivalence principle can also be stated as "the effects of gravity are exactly equivalent to the effects of acceleration."

Form the given option the correct option which can be filled in the blank is option 2 which states that the effects of gravity are exactly equivalent to the effects of acceleration.

Thus Einstein's equivalence principle says that  the effects of gravity are exactly equivalent to the effects of acceleration.

Learn more about the Einstein's equivalence principle here

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A violin has a string of length
0.320 m, and transmits waves at
622 m/s. At what frequency does
it oscillate?

Answers

Answer:

1.9kHz

Explanation:

Given data

wavelength [tex]\lambda= 0.32m[/tex]

velocity [tex]v= 622 m/s[/tex]

We know that

[tex]v= f* \lambda\\\\f= v/ \lambda[/tex]

substitute

[tex]f= 622/ 0.32\\\\f= 1943.75\\\\f= 1.9kHz[/tex]

Hence the frequency is 1.9kHz

Answer:

971.2

Explanation:

It was right on acellus :)

If there is "waste" energy, does the Law of Conservation of Energy still apply? ​

Answers

Explanation:

Yes, the law of conservation of energy still applies even if there is waste energy.

The waste energy are the transformation products of energy from one form to another.

According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".

But of then times, energy is lost as heat or sound within a system.

If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.

1.18. Which of the following is/are supplementary unit(s)? (1) Kelvin
(II) Newton (III) Second (IV) Radian
A. I and III only
C. I and II only
B. IV only
D. I, II and IV only

Answers

Answer:

B. IV only

Explanation:

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?

Answers

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

               d)  2.25 m

               e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 =  9.8t^2 / 2

hence d = 4.9t^2  ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t  ---- ( 2 )

equating equation 1 and 2

3 = 7.67t   therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec  = 1.92 m/sec

What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm

Answers

Solution :

1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).

2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).

3. Non-selective scatter takes place when particle size in greater than the wavelength  (λ).

We have the sizes of different particles :

[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]

Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]

Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]

Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]

Wavelength           [tex]$ O_2 $[/tex]         Smoke particles    Cloud droplets     Rain droplets

                            [tex]$10^{-10} \ m$[/tex]        [tex]$ 3 \times 10^{-7} \ m$[/tex]           [tex]$ 2 \times 10^{-5} \ m$[/tex]              [tex]$ 3 \times 10^{-3} \ m$[/tex]

[tex]$5500 \times 10^{-4} \ m$[/tex]      Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$11 \times 10^{-6} \ m $[/tex]         Rayleigh    Rayleigh            Non-selective      Non-selective

[tex]$1600 \times 10^{-10} \ m $[/tex]    Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$10^{-2} \ m $[/tex]                 Rayleigh      Rayleigh               Rayleigh          Mie

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