A 0.70-m radius cylindrical region contains a uniform electric field that is parallel to the axis and is increasing at the rate 5.0 × 1012 v/m?s. the magnetic field at a point 1.2 m from the axis has a magnitude of:

The speed of a proton is about 5.0 × 10⁵ m/s

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Let's recall the Kinetic Energy formula:

[tex]\boxed {E_k = \frac{1}{2}mv^2 }[/tex]

Ek = kinetic energy ( J )

m = mass of object ( kg )

v = speed of object ( m/s )

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

potential difference = ΔV = -1300 V

mass of proton = m = 1.67 × 10⁻²⁷ kilograms

charge of proton = q = 1.60 x 10⁻¹⁹ coulombs

initial speed of proton = v₁ = 0 m/s

Asked:

final speed of proton = v = ?

Solution:

[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]

[tex]qV_1 + \frac{1}{2}mv_1^2 = qV_2 + \frac{1}{2}mv_2^2[/tex]

[tex]q(V_1 - V_2 ) = \frac{1}{2}m( v_2^2 - v_1^2 )[/tex]

[tex]q( 0 - V ) = \frac{1}{2}m ( v^2 - 0^2 )[/tex]

[tex]-q \Delta V = \frac{1}{2} m v^2[/tex]

[tex]v^2 = -2mq \Delta V[/tex]

[tex]v = \sqrt { (-2q \Delta V) \div m }[/tex]

[tex]v = \sqrt { -2 \times 1.60 \times 10^{-19} \times (-1300) \div (1.67 \times 10^{-27})}[/tex]

[tex]v \approx 5.0 \times 10^5 \texttt{ m/s}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Mathematics

Chapter: Energy

The speed of a proton that has been accelerated from rest through a potential difference of - 1300 v  would be 4.99 × 10⁵ meters/ second.

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge.

As given in the problem we have to find out the speed  of a proton that has been accelerated from rest through a potential difference of -1300 volts,

The speed of the proton can be calculated with the expression given as follows,

1/2 × mass ×velocity ² = electric charge × potential difference  

v² = 2qV /m

v = √(2qV/m)

  = √ ( 2×1.6×10⁻¹⁹ ×1300 / 1.67×10⁻²⁷)

  =4.99 × 10⁵ meters/seconds

Thus, the speed of the proton would be 4.99 × 10 ⁵  meters/seconds.

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The work done by the gas is about 30 L.atm

[tex]\texttt{ }[/tex]

The Ideal Gas Law and Work formula that needs to be recalled is:

[tex]\boxed {PV = nRT}[/tex]

[tex]\boxed { W = P \Delta V }[/tex]

where:

W = Work ( J )

P = Pressure ( Pa )

V = Volume ( m³ )

n = number of moles ( moles )

R = Gas Constant ( 8.314 J/mol K )

T = Absolute Temperature ( K )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial volume = Vo = 15 L

final volume = V = 35 L

constant pressure = P = 1.5 atm

Asked:

word done by the gas = W = ?

Solution:

[tex]W = P \Delta V[/tex]

[tex]W = P \times ( V - Vo )[/tex]

[tex]W = 1.5 \times ( 35 - 15 )[/tex]

[tex]W = 1.5 \times 20[/tex]

[tex]W = 30 \texttt{ L.atm }[/tex]

[tex]\texttt{ }[/tex]

The work done by the gas is about 30 L.atm

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

The work energy, w, gained or lost by the system when the gas expands is 30 Joules.

Given the following data:

To find the work energy, w, gained or lost by the system when the gas expands:

Mathematically, the work energy by a system is given by the formula:

[tex]Work = P \delta V[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Work\;energy = 1.5(35-15)\\\\Work\;energy = 1.5(20)[/tex]

Work energy = 30 Joules.

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The magnitude of the velocity of the water particle is approximately 2.6 m/s, and the magnitude of the acceleration of the water particle is approximately 7.5 m/s^2.

The magnitudes of the velocity v and acceleration a of a water particle can be calculated using the concepts of angular velocity, and the relationships between linear and angular motion.

The velocity of the water particle v has two components: a tangential component due to the rotation (equal to the angular velocity times the distance from the center of rotation), and a radial component due to the flow of water along the tube.

The tangential velocity Vt = ωr, where r = 0.85 m is the distance of the water particle from the center of rotation and ω = 2.2 rad/s is the angular velocity. Substituting these values we find that Vt = 2.2 rad/s * 0.85 m = 1.87 m/s. The radial velocity Vr = l_dot = 1.8 m/s is given in the problem.

The total velocity is obtained by combining the radial and tangential velocities. Its magnitude is given by v = sqrt(Vr^2 + Vt^2). Substituting the known values, we find that the magnitude of the velocity v of the water particle is approximately 2.6 m/s.

The acceleration of the particle also has two components: a radial component equal to the square of the tangential velocity divided by the radius, and a tangential component equal to the product of the angular velocity and the radial velocity. Therefore, the magnitude of the acceleration a is given by a = sqrt((Vt^2 / r)^2 + (ωVr)^2), which gives us a value of approximately 7.5 m/s^2.

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The magnitude of the velocity [tex]v[/tex] of the water particle is approximately 2.6 m/s. The magnitude of the acceleration [tex]a[/tex] of the water particle is 4.114 m/s².

Velocity due to rotation (tangential velocity):

The tangential velocity [tex]v_{tangential}[/tex] is given by

[tex]v_{tangential} = l \cdot \dot{\theta}[/tex]

Substituting the values:

[tex]v_{tangential} = 0.85 \text{ m} \cdot 2.2 \text{ rad/s} = 1.87 \text{ m/s}[/tex]

Velocity along the tube:

The velocity along the tube [tex]v_{tube}[/tex] is given by [tex]\dot{l}[/tex], which is already provided as 1.8 m/s.

Resultant velocity:

The total velocity of the particle [tex]v[/tex] is the vector sum of the tangential and tube velocities. Since the angle [tex]\beta[/tex] between these two components is given, we can use the Pythagorean theorem:

[tex]v = \sqrt{( v_{tangential} )^2 + ( v_{tube} )^2}[/tex]

Substituting the values:

[tex]v = \sqrt{(1.87 \text{ m/s})^2 + (1.8 \text{ m/s})^2} \approx 2.6 \text{ m/s}[/tex]

Acceleration

Centripetal acceleration:

The centripetal (radial) acceleration [tex]a_c[/tex] is given by:

[tex]a_c = l \cdot \dot{\theta}^2[/tex]

Substituting the values:

[tex]a_c = 0.85 \text{ m} \cdot (2.2 \text{ rad/s})^2 = 4.114 \text{ m/s}^2[/tex]

Tangential acceleration:

The tangential acceleration [tex]a_t[/tex] will be zero because the angular velocity is constant (no angular acceleration).

Total acceleration:

The total acceleration [tex]a[/tex] is due to the centripetal acceleration alone, as there is no tangential component. Hence:

[tex]a = a_c = 4.114 \text{ m/s}^2[/tex]

Ans: The source Address field specifies the station that sends the frame. 

The format of an Ethernet frame includes a destination address at the beginning which contains the address of the device which is sending the frame. And, the source address tells us which station the information is received from. 

Answer:

In the field of radioactivity, the half-life is usually defined as the time required by an unstable radioactive isotope to disintegrate half of its initial composition. For different radioactive isotope elements, this value of half-life is different.

For example, the half-life of uranium-238 is approximately 4.5 billion years and the half-life of Carbon-14 is nearly 5700 years.

During the time of one half-life of a radioactive isotope, half of the parent atoms are disintegrated and forms a comparatively stable daughter isotope. This means that half of the initial concentration of the unstable isotope is reduced.

Answer:

The foot push of the cyclist exerted a force of 63 Newton.

Explanation:

Given data

We can find the force exerted by the foot push of the cyclist using the following expression.

P = F × v

v = P/F = 500 W/ (8.0 m/s) = 63 N

The foot push of the cyclist exerted a force of 63 Newton.

Answer:

static friction

Explanation:

she's not moving so she has static friction

Answer:

[tex]v_s= 37.8 m/s[/tex]

Explanation:

As per Doppler's effect when source and observer moves relative to each other then the frequency of the sound observed is different from the real frequency

When source is moving towards the stationary observer then we have

[tex]f_1 = f_o(\frac{v}{v - v_s})[/tex]

now when source of sound moving away from stationary observer then we have

[tex]f_2 = f_o(\frac{v}{v + v_s}[/tex]

now from above two equations

[tex]\frac{f_1}{f_2} = \frac{v + v_s}{v - v_s}[/tex]

here we know that

[tex]f_1 = 1000 hz[/tex]

[tex]f_2 = 800 hz[/tex]

v = 340 m/s

now we have

[tex]\frac{1000}{800} = \frac{340 + v_s}{340 - v_s}[/tex]

[tex]5(340 - v_s) = 4(340 + v_s)[/tex]

[tex]340 = 9 v_s[/tex]

[tex]v_s = 37.8 m/s[/tex]

The final frequency is 453.16 Hz when the tension is tripled.

We know that:

f = v / λ

Where,

f = frequency   of sound wave.

v= speed of sound wave

λ = wavelength of sound wave

So, for a fixed wavelength ( let it be λ), frequency is directly proportional to velocity of the sound wave.

Again for a certain tension T, velocity of sound wave in a string is given by, v = √(T/μ) where, μ is mass per unit length of the string.

So, when this tension is tripled; velocity of sound wave becomes √3 times of its initial value and as frequency is directly proportional to the velocity,  final frequency becomes = √3 × initial frequency

= √3 × 261.63 Hz. ( as given in the question, initial frequency = 261.63 Hz)

= 453.16 Hz.

Hence,  when this tension is tripled, the frequency will be 453.16 Hz.

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The concept Ohm's law is used here to determine the resistance. The resistance is found to be 16 ohm .The correct option is A.

The relationship between the electric current and the potential difference is given by the Ohm's law. The current which flows through the conductors is directly proportional to the voltage applied. Mathematically the relationship is given as:

V = IR

V - Potential difference

R - Resistance

I - Current

R = V / I

R = 120 / 7.6

R = 15.7 ohm ≈ 16 ohm

The ohm's law holds true if the provided temperature and the other physical factors remain constant. In certain components, increasing the current raises the temperature. In this case Ohm's law is violated.

It is the Ohm's law which maintains the desired voltage drop across the electronic components. It helps to determine the voltage, resistance or current of an electric circuit.

Thus the correct option is A.

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The volume of a unit cell of lead, which is a face-centered cubic, can be determined by calculating the edge length of the cell using its atomic radius and then cubing the edge length. The volume of a lead unit cell with an atomic radius of 0.175 nm is 1.205 * 10^-29 m^3.

To calculate the volume of a unit cell of lead, we need to understand that lead crystalizes in a face-centered cubic lattice, where the edge length of the unit cell equals the square root of 2 times the atomic radius, doubled (since the diameter is 2 times the radius). So we first calculate the edge length, a = 2 * atomic radius * sqrt(2), and then calculate the volume of the unit cell, which is a cubic volume, V = a^3.

Given the atomic radius of lead is 0.175 nm or 0.175 * 10^-9 meters, we calculate:

Edge length, a = 2 * 0.175 * 10^-9 meters * sqrt(2) = 0.494 * 10^-9 meters.

The volume V = (0.494 * 10^-9 meters)^3 = 0.1205 * 10^-27 m^3 or 1.205 * 10^-29 m^3.

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A solar eclipse occurs when the Moon blocks the Sun from the Earth's viewpoint during a New Moon phase. The Moon also needs to be crossing the line of nodes, the intersection of the Earth's and Moon's orbit plain.

A solar eclipse occurs when the Moon moves between the Sun and the Earth, casting a shadow on the Earth. This event happens only during a New Moonphase, which is one of the conditions needed. Not all New Moon phases result in a solar eclipse because the Moon's orbit is tilted relative to the Earth's orbit around the Sun. Therefore, for a solar eclipse to take place, the Moon must be in the New Moon phase, and it must also be at a position in its orbit where it crosses the plane of the Earth's orbit around the Sun. This place is known as the line of nodes.

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Neglecting air resistance, the rock would have fallen a distance of 20 meters after 2 seconds, calculated using the formula d = gt²/2 with g as the acceleration due to gravity (10 m/s²).

If you drop a rock from a tall building neglecting air resistance, the distance it has fallen after 2 seconds can be calculated using the formula for the distance fallen under constant acceleration, which is d = gt²/2, where g is the acceleration due to gravity (approximately 10 m/s²) and t is the time in seconds the object has been falling.

Plugging in the values, we get:

d = 10 m/s² × (2 s)² / 2 = 20 m

Therefore, the rock would have fallen 20 meters after 2 seconds.

Imagine that you're standing in a large room when a loud noise is made. You begin hearing a series of echoes. The characteristic of sound that best explains what just happened is: Reverberation.

707 N

185×0.39×9.8

I just did it and found the answer