In the figure conducting rod is shown moving in a uniform magnetic field. After the free charges in the bar have stopped moving up or down the bar; which of the following statements are true? Check all that apply: Supporting Material X X X X X X X X X X X X X An electric field is established in the rod directed from point a to point b The magnitude ofthe potential difference in the rod is inversely proportional to the length Loftherod The magnitude ofthe potential difference in the rod is proportional to the strength of the magnetic field: The magnitude of the potential difference in the rod is proportional to the velocity v of therod The free charges in the rod are acted upon by a magnetic force:

Answer:

(i) Please see graph of the motion of the particles created with MS Excel and the calculations in the following sections

(ii) The time of collision is approximately 1.0 seconds

(iii) The common horizontal distance of point collision from the point of projection is approximately 25.2 meters

Explanation:

The velocity with which the first projectile was fired, v₁ = 21 m/s

The angle to the horizontal the particle is launched = 53.1°

The time at which the other particle was launched = 1 second later

The location from which the other particle was projected = 0.3 m below the first particle

The initial velocity of the second particle = 31.5 m/s

The angle to the horizontal at which the second particle was projected, θ = 36.9°

(i) The height reached, by each of the particle is given as follows;

y = u·t - 1/2·g·t²

For the first projectile, we have;

y = 21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3

For the second projectile, we have;

y= 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

If the two projectiles collide, we get;

21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3  = 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

Using a graphing calculator for simplifying, we get;

-11.93·t₁ + 12.2 = 0

t₁ = 12.2/11.93 ≈ 1.02

Therefore, at time t₁ = 1.02 seconds, after the launch of the second particle, the two particle will be at the same vertical height

However, whereby at the time, t₂, the particles collide, the horizontal distance travelled, 'x', will be equal;

We have;

x = u·cos(θ)·t₁

For the first particle, we have;

x₁₁ = 21 × cos (53.1°) × (t₂ + 1)

For the second particle, we have;

x₂₂ = 31.5 × cos (36.9°) × t₂

At the point of collision, we have;

x₁ = x₂

∴ 21 × cos (53.1°) × (t₂ + 1) = 31.5 × cos (36.9°) × t₂

31.5 × cos (36.9°) × t₂ - 21 × cos (53.1°) × t₂ = 21 × cos (53.1°)

t₂ = 21 × cos (53.1°)/(31.5 × cos (36.9°)  - 21 × cos (53.1°) ) = 1.00219236871

t₂ ≈ 1.0 seconds

Given that t₁ ≈ t₂, the particles reach the same height and the same horizontal distance at the same time, t₂ ≈ 1.0 and therefore, they collide.

(ii) The time of collision is found above as t₁ ≈ t₂ ≈ 1.0 seconds

(iii) The horizontal distance of the point of collision from the starting point, 'x', is given as follows;

x = 21 × cos (53.1°) × (1.0 + 1) ≈ 25.2

The horizontal distance of the point of collision from the starting point, x ≈ 25.2 meters

The vertical distance of the point of collision from the starting point of the second particle, 'y', is given as follows;

y = 21 × (1+1)×sin(53.1°) - 9.81 × (1+1)²/2 + 0.3 ≈ 14

The vertical distance of the point of collision from the starting point of the second particle, y ≈ 14 meters

The magnitude of the distance from the starting point of the second particle, r = √(25.2² + 14²) ≈ 28.8

The magnitude of the distance from the starting point of the second particle, r ≈ 28.4 meters.

Answer:

The height of the image of the candle is 20 cm.

Explanation:

Given that,

Size of the candle, h = 12 cm

Object distance from the candle, u = -6 cm

Focal length of converging lens, f = 15 cm

To find,

The height of the image of the candle.

Solution,

Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.

\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\)

v is image distance

\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm\)

If h' is the height of the image. Magnification is given by :

\(m=\dfrac{h'}{h}=\dfrac{v}{u}\)

\(h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm\)

So, the height of the image of the candle is 20 cm.

disadvantages

Explanation:

releases heat when not required

opposes motion

Answer:

0.500 T

Explanation:

Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.

Explanation:

yan lng po Alam ko

Answer:

Range = 9

Explanation:

From the question,

Mean = summation data/Total frequency

Mean = (6+7+x+13+10)/5

If the mean is 8,

8 = (36+x)/5

Solve for x

36+x = 8×5

36+x = 40

x = 40-36

x = 4.

Range is the difference between the biggest and the smallest value in a given distribution.

from the above,

Range = 13-4

Range = 9

The moment of inertia of the rider, box, and rocket about the pole is 170 kg * (6 m)^2 = 6120 kg · m^2.  The initial angular acceleration of the rider is 0.098 rad/s^2.

A) The moment of inertia of the rider, box, and rocket about the pole can be calculated using the formula for the moment of inertia of a point mass rotating around an axis. The moment of inertia (I) is given by the product of the mass (m) and the square of the distance (r) from the axis of rotation:

I = m * r^2

Since the mass of the rider, box, and rocket is given as 170 kg and the rod is neglected, the moment of inertia will be the same for all components. Therefore, the moment of inertia of the rider, box, and rocket about the pole is 170 kg * (6 m)^2 = 6120 kg · m^2.

B) The initial angular acceleration of the rider can be determined using Newton's second law of rotational motion, which states that the torque (τ) is equal to the moment of inertia (I) multiplied by the angular acceleration (α). The torque can be calculated as the product of the force (F) applied perpendicular to the path of the rider and the distance (r) from the axis of rotation:

τ = F * r

Rearranging the formula and substituting the given values:

α = τ / I = (98 N * 6 m) / 6120 kg · m^2 = 0.098 rad/s^2

Therefore, the initial angular acceleration of the rider is 0.098 rad/s^2.

C) The relationship between angular velocity (ω) and time (t) is given by the equation ω = α * t. Rearranging the formula to solve for time:

t = ω / α

The rider's velocity can be converted to angular velocity using the formula v = r * ω. Rearranging the formula to solve for angular velocity:

ω = v / r = 5 m/s / 6 m = 0.8333 rad/s

Substituting the values into the equation, we get:

t = 0.8333 rad/s / 0.098 rad/s^2 = 8.5 s

Therefore, the time required for the rider's velocity to reach 5 m/s is 8.5 seconds.

D) To find the mass per second that must leave the rocket to develop the given thrust (F), we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (dp/dt). In this case, the force is equal to the thrust of the rocket, and the change in momentum is the mass per second leaving the rocket (dm/dt) multiplied by the exit velocity (v):

F = dm/dt * v

Rearranging the formula and substituting the given values:

dm/dt = F / v = 98 N / 390 m/s = 0.2513 kg/s

Therefore, the mass per second that must leave the rocket to develop the thrust of 98 N is 0.2513 kg/s.

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True. The field near a long straight wire carrying a current is directly proportional to the distance from the wire and inversely proportional to the current flowing through the wire. This is known as the Biot-Savart law.

The field lines around the wire form circles perpendicular to the wire, and the direction of the field can be determined using the right-hand rule. The strength of the field decreases as the distance from the wire increases, and the strength of the field also decreases as the current flowing through the wire decreases. Therefore, the statement that the field near a long straight wire carrying a current is inversely proportional to the current flowing through the wire is true.

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Answer: everyone has a religious belief

Explanation:

You have a belief as well called Scientology

You believe science is more powerful then religion

They also need it as an excuse to think there is somewhere to go when you die so they make the good place heaven and hell

Answer:

people need to believe that there is a higher power guiding them. to explain events that they don't understand. that there is a reason for their existence. people need something to believe in. and you can't say someone's belief is wrong just because you don't agree with it or share that same belief.  

Explanation:

Answer:

acceleration = 6.29 ft^2/s

distance traveled by car = 405 ft

Explanation:

formula 1 : u = v - (a*t)

where u = initial velocity

v = final velocity

a = acceleration

t = time.

solving for acceleration.

24 = 58 - (a*5.4)

a = (58 - 24)/ 5.4

a = 340/ 54 = 6.29 ft^2/s

formula 2: v = d/t - (a*t/2)

58 = d/5.4 - ((340*54)/20)

d = 405 ft

It is all about the dilation of pupils. When darkness falls, your pupils enlarge, whereas in daylight, your pupils shrink.

I am joyous to assist you at any time.

Answer:

106.8

Explanation:

If you multiply 6 and 17.8, you'll get 106.8.

If that's wrong, my apology's !

Answer:

fdsbgdfshtrsnbfgsbnhgr

Explanation:

cdfrgresgtrshtrwhtrwhtwjhdgngdabfeahydrtgnb

The  centripetal force the passenger experience is 30167.52 N

The centripetal force experienced by the passenger can be calculated using the formula:

Fc = (m)(r)(ω^2)

where Fc is the centripetal force, m is the mass of the passenger, r is the radius of the loop, and ω is the angular speed of the roller coaster.

Given:

m = 80 kg

r = 12 m

ω = 3.14 rad/s

Substituting these values into the formula, we get:

Fc = (80 kg)(12 m)(3.14 rad/s)^2

Fc = 30167.52 N

Therefore, the passenger experiences a centripetal force of approximately 30167.52 N when the roller coaster reaches an angular speed of 3.14 rad/s.

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Answer:

the sun is currently a mian sequence star

and will remain so for another 4-5 billion

Answer:

bc we play when we are bored and some people use it as a tradition to play on thanksgiving

Answer:

1/3

Explanation:

Velocity ratio = distance moved by effort / distance moved by load

Answer:

The net charge on each sphere is -1.5 q

Explanation:

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

\(Q = \dfrac{Q_1 + Q_2}{2}\)

Therefore, by substituting the known values of the variables, we have;

\(Q = \dfrac{5 \ q+ (-8 \ q)}{2} = -\dfrac{3 \ q}{2} = -1.5 \ q\)

The net charge on each sphere, Q = -1.5 q.

The net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

Conductors are materials that allow the electrons which are the carriers of the charges to move between them, and when two conductors come in contact, the available charge is shared by the two conductors and the resultant like charges will spread on the surface of the conductor due to the repellent effect between similar charges such that if the conductors are identical, the resultant charge becomes evenly shared by the conductors when they become separated again

The given parameters of the conducting spheres meant to touch are;

The net charge on the first sphere, Q₁ = +5.0 q

The net charge on the second sphere, Q₂ = -8.0 q

The net charge on each sphere after touching and then separated, 'Q', is given as follows;

\(Q=\dfrac{Q_1+Q_2}{2}\)

Therefore, by substituting the known values of the variables, we have;

\(\dfrac{5q+(-8q)}{2}=-1.5q\)

Hence the net charge on each sphere after spheres are brought together, allowed to touch is -1.5 q

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Answer:

Velocity of the car at the bottom of the slope: approximately \(20.3\; \rm m \cdot s^{-2}\).

It would take approximately \(3.9\; \rm s\) for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

\(v^2 - u^2 = 2\, a\cdot x\).

Given:

Rearrange the equation \(v^2 - u^2 = 2\, a\cdot x\) and solve for \(v\):

\(\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}\).

Calculate the time required for reaching this speed from \(u = 3\; \rm m \cdot s^{-1}\) at \(a = 4.5\; \rm m \cdot s^{-2}\):

\(\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}\).

The statements which describe some of the original ideas proposed about the universe during the big bang include:

This is the theory about the formation of the universe which states that a single point began to expand which resulted in its formation.

This was as a result of observations and ideas which can be seen mentioned above.

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Answer:

B. The universe was opaque.

D. The universe was very hot and dense.

E. The universe expanded from a single point.

Explanation:

Answer:

4 x the frequncy

Explanation:

So 4 x the number

The velocity of the ball as it exits the sand is 6m/s.

When the ball rolls into the sand, it experiences a force of friction acting against its motion, which causes it to slow down. The amount of frictional force depends on the properties of the sand and the ball's velocity. Assuming that the ball rolls horizontally into the sand and comes out horizontally as well, the conservation of momentum applies, which means that the momentum of the ball before it enters the sand is equal to the momentum of the ball after it exits the sand.

We can use the equation for conservation of momentum to calculate the final velocity of the ball:

Initial momentum = Final momentum

mv1 = mv2

where m is the mass of the ball, v1 is the initial velocity of the ball, and v2 is the final velocity of the ball.

Substituting the given values, we get:

2 kg x 6 m/s = 2 kg x v2

12 kg m/s = 2 kg x v2

v2 = 6 m/s

Therefore, the final velocity of the ball as it exits the sand is 6 m/s.

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radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

Answer:

sublimation and evaporation

Explanation:

evaporation is from liquid to gas and sublimation is solid to liquid, as water vapor is a gas both of these methods could be used

Answer:

A pure substance in the gaseous state contains more energy than in the liquid state, which in turn contains more energy than in the solid state. Particles has the highest kinetic energy when they are in the gaseous state.

Explanation:

Sana makatulong.

Answer:

Impulse

Explanation:

Impulse is force times time

(i)Therefore, it takes approximately 9.27 hours to reach its full power output.(ii)It is necessary to have quick-start power sources, this helps maintain a stable and reliable electricity supply even when wind speeds fluctuate.(c)The Bremsstrahlung effect needs to be considered to ensure proper radiation protection.(d) The near-field region is characterized by strong electric and magnetic fields while the far-field region represents the radiation zone.

(i) To calculate the time it takes for the power station to reach its full power output, we can use the formula:

Energy = Power × Time

Given that the power station generates 6120 MWh of energy over a 10-hour period and the rated power at full capacity is 660 MW, we can rearrange the formula to solve for time:

Time = Energy ÷ Power

Converting the energy to watt-hours (Wh):

Energy = 6120 MWh × 1,000,000 Wh/MWh = 6,120,000,000 Wh

Converting the power to watt-hours (Wh):

Power = 660 MW × 1,000,000 Wh/MW = 660,000,000 Wh

Now we can calculate the time:

Time = 6,120,000,000 Wh ÷ 660,000,000 Wh ≈ 9.27 hours

Therefore, it takes approximately 9.27 hours (or 9 hours and 16 minutes) for the power station to reach its full power output.

(ii) The type of power station that can be started up fastest is a gas-fired power station. Gas-fired power stations can reach full power output relatively quickly because they use natural gas combustion to produce energy.

In contrast, other types of power stations, such as coal-fired or nuclear power stations, have longer start-up times. Coal-fired power stations require time to heat up the boiler and generate steam, while nuclear power stations need to go through a complex series of procedures to ensure safe and controlled nuclear reactions.

This is relevant to optimizing the usage of windfarms because wind power is intermittent and dependent on the availability of wind. This helps maintain a stable and reliable electricity supply even when wind speeds fluctuate.

(c) The Bremsstrahlung effect is a phenomenon that occurs when charged particles, such as electrons, are decelerated or deflected by the electric fields of atomic nuclei or other charged particles. As a result, they emit electromagnetic radiation in the form of X-rays or gamma rays.

In shielding design, the Bremsstrahlung effect needs to be considered to ensure proper radiation protection. These materials effectively absorb and attenuate the emitted X-rays and gamma rays, reducing the exposure of individuals to harmful radiation.

(d) The electromagnetic field output from an antenna can be represented by two main regions:

Near-field region: This region is closest to the antenna and is also known as the reactive near-field. It extends from the antenna's surface up to a distance typically equal to one wavelength. In the near-field region, the electromagnetic field is characterized by strong electric and magnetic field components.

Far-field region: Also known as the radiating or the Fraunhofer region, this region extends beyond the near-field region.The electric and magnetic fields are perpendicular to each other and to the direction of propagation.  The far-field region is further divided into the "Fresnel region," which is closer to the antenna and has some characteristics of the near field, and the "Fraunhofer region," which is farther away and exhibits the properties of the far-field.

The transition between the near-field and the far-field regions is gradual and depends on the antenna's size and operating frequency. The size of the antenna and the distance from it determine the boundary between these regions.

In summary, the near-field region is characterized by strong electric and magnetic fields, while the far-field region represents the radiation zone where the energy is radiated away as electromagnetic waves.

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To find the spherical convex mirror radius, learn the focal length and magnitude.

What is focal length?

The focal length of a thin lens in air is measured from the lens's center to its primary foci, also known as focal points. The focal length of a converging lens, such as a convex lens, is positive and determines how far a collimated light beam must travel to focus on a single point.

What is magnitude?

Magnitude typically refers to size or distance. By comparing the size and motion speed of the object, we can relate magnitude to movement. The object's or quantity's size determines its magnitude.

R=0.123m

Focal length E= R/2

E= - 0.0615m.

Therefore, the focal length is 0.0615m.

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Moving a magnet in and out of a coil will make the needle on the galvanometer move to either side. This phenomenon can be explained in terms of magnetic induction. When the magnet is moved into the coil, the magnetic field lines of the magnet cut across the coil's wire.

This cutting of magnetic field lines induces an electric current in the coil, according to Faraday's law of electromagnetic induction. This induced current flows through the galvanometer, causing the needle to move. The direction of the needle's movement depends on the direction of the current.

When the magnet is moved out of the coil, the magnetic field lines again cut across the coil's wire, inducing a current in the opposite direction. This causes the needle on the galvanometer to move in the opposite direction. In summary, the movement of the magnet in and out of the coil induces an electric current in the coil, which is detected by the galvanometer and causes the needle to move.

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