considering fig.29 and the equation for vr(t), what is the approximate phase of vr(t) in degree for vo

Electromagnetic waves are a form of radiation that occurs when electric fields and magnetic fields are coupled with each other. Thus, option C is correct.

An electromagnetic field is formed with an electric field and a magnetic field that are dependent on each other. The two fields are perpendicular to each other and travel in that direction as well.

Electromagnetic waves are not elastic waves either and can travel in a vacuum. They are different from mechanical waves as electromagnetic waves can travel in the absence of a medium. They travel at the speed of 3x\(10^{8}\)m/s. Thus, only option C is right.

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Complete question:-

Which of the following are the characteristics of electromagnetic waves?

1. They are elastic waves.

2. They can also move in a vacuum.

3, They have electric and magnetic components which are mutually perpendicular.

4, They move with a speed equal to 3 lakh meters per second.

Select the correct answer using the code given below :

A) 1, 2, 3 and 4

B) 1, 2, and 4

C) 2 and 3 only

D) 3 and 4 only

The radioactive decay is shown in the image attached to this answer

Answer:

Alpha

Explanation:

An alpha decay is said to have occurred when the mass number of the daughter nucleus decreases by four units and the atomic number of the daughter nucleus decreases by two units.

If we look at the equation from left to right, we will notice that the mass number of the daughter nucleus decreased by four units and the atomic of the daughter nucleus number by two units compared to that of the parent nucleus leading to the emission of an alpha particle.

Answer:

divide the number of times the event occurs by the length of time.

Explanation:

Work is done on an object by a force of 10.0 N pushing it 5.00 m is 50.0 joule.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.

Given parameters:

Force applied on the object is: F = 10.0 N.

Displacement of the object is: d = 5.00 meter.

Hence, Work is done on the object = force × displacement

= 10. N × 5.00 meter

= 50.0 Joule.

So, Work is done on an object by a force of 10.0 N pushing it 5.00 m is 50.0 joule.

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The speed of sound in the fluid is 1.788 x 10³ m/s.

The speed of sound in a fluid can be calculated using the formula:
v = √(B/ρ)
Where v is the speed of sound, B is the bulk modulus, and ρ is the density.

The speed of sound refers to the rate at which sound waves travel through a medium while the bulk modulus is a measure of a substance's resistance to compression under pressure. On the other hand, density is a measure of the amount of mass in a given volume of a substance. It is defined as the mass of a substance per unit volume.

Plugging in the given values, we get:
v = √((3.2 x 10⁹ Pa)/(1.0 x 10³ kg/m³))

Simplifying the equation, we get:
v = √(3.2 x 10⁶ m²/s²)

Taking the square root, we get:
v = 1.788 x 10³ m/s

Therefore, in the given fluid, the speed of sound is 1.788 x 10³ m/s.

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Answer:

2.61 \(\frac{m}{s}\)

Explanation:

v= d/t

v = 6m/2.3 s

v=2.61 \(\frac{m}{s}\)

Answer:

\(\boxed {\tt v\approx2.61 \ m/s}\)

Explanation:

Velocity can be found by dividing the displacement by the time.

\(v=\frac{d}{t}\)

The displacement is 6 meters and the time is 2.3 seconds.

\(d= 6 \ m\\t= 2.3 \ s\)

Substitute the values into the formula.

\(v=\frac{6 \ m}{2.3 \ s}\)

Divide.

\(v=2.60869565 \ m/s\)

Let's round to the nearest hundredth. The 8 in the thousandth place tells us to round the 0 to a 1.

\(v\approx2.61 \ m/s\)

The velocity of the marble is about 2.61 meters per second.

The EMF of a cell represents its maximum potential difference, while the potential difference across the cell when passing a current through an external resistance is reduced due to the internal resistance of the cell.

The electromotive force (EMF) of a cell refers to the maximum potential difference that the cell can provide when it is not connected to any external circuit. It is essentially the voltage produced by the chemical reactions occurring within the cell. On the other hand, the potential difference across the cell when it is passing a current through an external resistance is often lower than the EMF.

This difference in values can be attributed to the internal resistance of the cell. When a current is drawn from the cell, it encounters resistance within the cell itself due to factors such as the resistance of the electrolyte and the internal structure of the cell. This internal resistance causes a voltage drop within the cell, reducing the potential difference across the external circuit.

According to Ohm's Law (V = I * R), when current flows through a resistance (R), a voltage drop (V) occurs. In this case, the internal resistance of the cell causes a voltage drop within the cell itself. Hence, the potential difference across the external resistance is slightly lower than the EMF of the cell.

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To reduce the intensity of the transmitted beam by a factor of 3.40, the polarizing disk should be rotated by an angle of approximately 30.6 degrees.

To reduce the intensity of the transmitted beam by a factor of 4.40, the polarizing disk should be rotated by an angle of approximately 36.4 degrees. To reduce the intensity of the transmitted beam by a factor of 11.0, the polarizing disk should be rotated by an angle of approximately 68.3 degrees.
The angle of rotation needed to achieve these reductions in intensity is dependent on the direction of the incident light and the transmission axis of the polarizing disk. When the two are parallel, as in this case, the angle of rotation needed is determined by the equation:
θ = arccos(sqrt(I'/I))
where θ is the angle of rotation, I is the initial intensity, and I' is the reduced intensity.

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Explanation:

Assuming STP

EACH 22.4 liters is a mole of H2

320 m^3 = 320 000 liters

320 000 liters / ( 22.4 liters/mole) = 14285.7 moles

From periodic table , each mole of H2 is 2.016 gm

18285. 7 moles * 2.016 gm/ mole = 28799.97 gm = ~ 28.8 kg

Answer:

we see to make things visible . if we don't see then we can not give the scene around us. on the way if we close our eyes will we can we home from different types of vehicles .

Explanation:

thank you Mark as a brainlist

The potential function and the gravitational field intensity function inside and outside of a thin ring with radius R, in terms of r (the distance from the center of the ring to the field point), are as follows:

Inside the ring (r < R): The potential function is given by V = -GM/r, and the gravitational field intensity function is E = GM/r², where G is the gravitational constant and M is the mass of the ring.

Outside the ring (r > R): The potential function is V = -GM/r + GM/R, and the gravitational field intensity function is E = GM/r², where G is the gravitational constant, M is the mass of the ring, and R is the radius of the ring.

Inside the ring (r < R), the potential function is given by V = -GM/r. This represents the gravitational potential due to the mass of the ring at a point inside the ring. The negative sign indicates that the potential decreases as the distance from the center of the ring decreases. The gravitational field intensity function is E = GM/r², representing the strength of the gravitational field at a point inside the ring. The field intensity decreases as the distance from the center of the ring increases, following an inverse square relationship.

Outside the ring (r > R), the potential function is V = -GM/r + GM/R. In addition to the potential due to the mass of the ring, there is an additional potential term GM/R, which arises from considering the ring as a point mass located at its center. The gravitational field intensity function remains the same as E = GM/r², indicating that outside the ring, the gravitational field follows an inverse square relationship with the distance from the center of the ring.

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Answer:

answer is 4

Explanation:

The charge that is moved is required.

The magnitude of charge moved is option 4. \(2\times 10^{-3}\ \text{C}\)

V = Potential difference = 10.0 V

W = Work done = \(2\times10^{-2}\ \text{J}\)

q = Magnitude of charge

When work done on moving a charge is divided by the potential difference we get the magnitude of the charge that is moved.

The charge of an object is the result of an excess or a lack of positive or negative charges on a body.

So, the magnitude of charge is given by

\(q=\dfrac{W}{V}\\\Rightarrow q=\dfrac{2\times 10^{-2}}{10}\\\Rightarrow q=2\times 10^{-3}\ \text{C}\)

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Answer:

I'm pretty sure it's c. I remember seeing a question like this and I think it was C

Explanation:

Answer:

70 km/h

Explanation:

70 divided by 2 = 35

Answer:

9 m/s

Explanation:

13.5 × \(\frac{2}{3}\) = 9

When an electrically neutral pith ball gains 4.0 * 10^23 electrons, its charge becomes negative, with a magnitude of approximately -1.6 * 10^-5 coulombs.

An electrically neutral object has an equal number of protons and electrons, resulting in a net charge of zero. However, when the pith ball gains electrons, the number of electrons exceeds the number of protons, giving the pith ball a negative charge.

Each electron has a charge of approximately -1.6 * 10^-19 coulombs, and gaining 4.0 * 10^23 electrons means the pith ball's charge will be approximately -6.4 * 10^-3 coulombs. Thus, the charge of the pith ball is q = -6.4 * 10^-3 C.

It's important to note that the charge of an object is quantized, meaning it can only exist in discrete multiples of the elementary charge (-1.6 * 10^-19 C). In this case, the pith ball gained a large number of electrons, resulting in a measurable negative charge.

The magnitude of the charge is determined by the number of excess electrons, while the negative sign indicates the presence of an excess of electrons compared to protons.

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Answer:

its a clause

Explanation:

because u have decided to go and go is a action

The gravitational force of a star on planet 2 orbiting it is half that of planet 1.

Newton's gravitational law formula for gravitational force between two objects

\(F = G \frac{M \: m}{r^2}\)

Given

Compare the gravitational force on two condition.

\(\frac{f1}{f2} = \frac{G \frac{M \: m_1}{r_1^2}}{G \frac{M \: m_2}{r_2^2}}\)

\(\frac{f1}{f2} = \frac{\frac{m_1}{r_1^2}}{\frac{m_2}{r_2^2}}\)

\(\frac{f1}{f2} = \frac{m_1 \: r_2^2}{m_2 \: r_1^2}\)

\(\frac{f1}{f2} = \frac{m_1 \: (2r_1)^2}{2m_1 \: r_1^2}\)

\(\frac{f1}{f2} = \frac{4r_1^2}{2r_1^2}\)

\(\frac{f1}{f2} = \frac{2}{1}\)

The ratio of gravitational force f1 : f2 = 2 : 1

2f2 = f1

\(f2 = \frac{f1}{2}\)

\(f2 = \frac{1}{2} f1\)

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Answer:

Schwarzschild radius is the radius of the boundary of a hole of this type.

Force(F) = 350 N

Work (W)= 0 J

Apply the formula:

Work = Force x Distance

Isolate distance(d)

Distance = Work/ Force

d = W/F = 0 /350 = 0 m

The distance that the object traveled is 0 (zero)

The graphs display the function solution y = ex and three numerical solutions -Euler, Improved Euler, and Runge-Kutta- at different intervals. The Relative Deviation on the right measures the difference between the numerical solutions and the exact function solution.

The analysis of each set of graphs involves comparing the numerical solutions (Euler, Improved Euler, and Runge-Kutta) with the exact function solution y = ex. The Relative Deviation provides an indication of how closely the numerical solutions approximate the exact solution.

To analyze each set of graphs, follow these steps:

By examining the graphs and assessing the Relative Deviation, one can analyze the accuracy and performance of the numerical solutions -Euler, Improved Euler, and Runge-Kutta- in approximating the exact function solution y = ex at different intervals or iterations.

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The work done in raising an object results in an increase in the object's gravitational potential energy.option C.

Gravitational potential energy is the energy possessed by an object due to its position relative to a reference point in a gravitational field. When an object is raised, work is done against the force of gravity, and the object gains potential energy as a result.To understand this, let's consider the formula for gravitational potential energy: PE = mgh, where PE represents potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or distance the object is raised.When the object is lifted to a higher position, h increases, resulting in an increase in potential energy. This increase in potential energy is equivalent to the work done on the object against gravity.

The object stores this energy, which can be released or converted into other forms of energy later.The other options, such as internal energy, kinetic energy, and heat energy, are not directly affected by the act of raising an object against gravity. Internal energy refers to the total energy of the object's microscopic particles, kinetic energy is associated with the object's motion, and heat energy is the transfer of energy due to temperature differences. In the context of raising an object, these forms of energy may not necessarily change unless there are other factors involved.option C.

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Answer:

2.39 °C

Explanation:

Specific heat of water = 4187 j / kg °C     (You eaither have to be given this, or know this  or look it up)

100 g of water = .1 kg of water

1000 j  / (.1 kg * °C) = 4187  j / (kg °C)

solve for C = 2.39°C

Answer:

The convex lens is shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens.

Explanation:

Answer:

Friction or tension

Explanation:

Friction stops an object from moving in the presence of force

The acceleration of each block is 1.225m/s^2

The string exerts 2.315N on 210g block, and 2.976N on 270g block

It will take 1.28s for each block to traverse

Part a

Acceleration, a of the block:

m1=270g

m2=210g

g=9.8

Acceleration=(m1-m2)*g/(m1+m2)

                    =(270-210)*9.8/(270+210)

                    =1.225m/s^2

Part b

Force that the string exerts on each block:

Force, T1 on smaller block:

T1=m2(g+a), where m2=0.21kg, g=9.8, and a=1.225m/s^2

T1=0.21(9.8+1.225)

T1=2.315N

Force exerted on smaller block=2.315N

Force, T2 on larger block

Here, m1=0.27kg, g=9.8, and a=1.225m/s^2

T2=0.27(9.8+1.225)

    =2.976N

Force exerted on larger block=2.976N

Part c

How long it will take each block to traverse 1.0m

h=1.0m

a=1.225m/s^2

\(t=\sqrt{\frac{2h}{a} }\)

\(t=\sqrt{\frac{2*1}{1.225} }\)

t=1.28s

The question was incomplete, complete question is given below:

A person holds a 210-g block that is connected to a 270-g block by a string going over a light pulley with no friction in the bearing (an Atwood machine). After the person releases the 210-g block, it starts moving upward and the heavier block descends. Part A: What is the acceleration of each block? Part B: What is the force that the string exerts on each block? Part C: How long will it take each block to traverse 1.0 m?

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your answer is c bro but if not then it's b

because you just choose c or b either one should work if you right it down right

Answer:

a) 0.60 kg cart has final velocity 3.0 m/s [E]

0.80 kg cart has final velocity 4.0 m/s [W]

b) 0.12 m

Explanation:

Take east to be positive.

a) Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.60) (-5.0) + (0.80) (2.0) = (0.60) v₁ + (0.80) v₂

-1.4 = 0.6 v₁ + 0.8 v₂

Kinetic energy is conserved in elastic collisions.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60) v₁² + (0.80) v₂²

18.2 = 0.6 v₁² + 0.8 v₂²

Solve the system of equations.

-1.4 = 0.6 v₁ + 0.8 v₂

-1.4 − 0.6 v₁ = 0.8 v₂

-1.75 − 0.75 v₁ = v₂

18.2 = 0.6 v₁² + 0.8 (-1.75 − 0.75 v₁)²

18.2 = 0.6 v₁² + 0.8 (3.0625 +2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 8 (3.0625 + 2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 24.5 + 21 v₁ + 4.5 v₁²

0 = 10.5 v₁² + 21 v₁ − 157.5

0 = v₁² + 2 v₁ − 15

0 = (v₁ − 3) (v₁ + 5)

v₁ = 3 or -5

Since u₁ = -5.0 m/s, v₁ must be 3.0 m/s.

Solving for v₂:

v₂ = -0.75 v₁ − 1.75

v₂ = -4.0 m/s

b) The compression of the spring is a maximum when the carts have the same velocity.

Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.60) (-5.0) + (0.80) (2.0) = (0.60 + 0.80) v

-1.4 = 1.4 v

v = -1.0

Energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ (m₁ + m₂) v² + ½ kx²

m₁u₁² + m₂u₂² = (m₁ + m₂) v² + kx²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60 + 0.80) (-1.0)² + (1200) x²

18.2 = 1.4 + 1200 x²

16.8 = 1200 x²

x² = 0.014

x = 0.12