Below is a schematic of a vapor power plant in which water steadily circulates through the four components. The water flows through the boiler and condenser at constant pressure and through the turbine and pump adiabatically. Kinetic and potential energy effects can be ignored. Process data follow: Process 4-1: constant pressure at 8 MPa from saturated liquid to saturated vapor Process 2-3: constant pressure at 8 kPa from x2

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

Answer:

23785m

Explanation:

V= 4780/9.9= 482.8km/h

d= v²/g = 482.8²/9.8= 23785m

Answer:

Explanation:

For pendulum , time period is

T = 2π √ l / g

T² = 4π² l/ g

l = / g T² / 4π²

= 9.8 x 5² / 4π²

= 6.21 m

Angular impulse = Torque x time

= 30 x 6.21 x .3 = increase in angular momentum

30 x 6.21 x .3 = I ω ; I is moment of inertia and ω is angular velocity .

I ω = 55.89

I = 1/3 m l²

= 1/3 x 5 x 6.21²

= 64.27

Angular kinetic energy = 1/2 I ω²

= 1/2(I ω)²/ I

= .5x 55.89²/ 64.27

= 24.30 J

Due to it , the centre of mass of pendulum increases by height h so that

h = l/2 ( 1 - cos θ )

1/2 I ω² = mg l/2 ( 1 - cos θ)

24.3 = 5 x 9.8 x 3.1 ( 1 - cos θ)

( 1 - cos θ) = .16

cos θ = .84

θ  = 33 degree.

The length of the pendulum is about 6.25m and the maximum angle of displacement would occur when the pendulum is kicked with the maximum force of 30N.

The length of a pendulum is determined by the formula T=2π√(L/g). Given that the period T is 5.0 seconds and g (acceleration due to gravity) is 9.8 m/s², we can rearrange the equation to solve for L (length), resulting in L = (T² * g) / (4π²) = (5.0)² * 9.8 / (4π²) ≈ 6.25m. This regard to the maximum angle of displacement, when a pendulum is in simple harmonic motion, which is typically for displacements less than 15 degrees, the displacement is directly proportional to its force. Therefore, the maximum angle would occur when the force is the greatest, in this case, when the student kicks the pendulum with a force of 30 N.

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Answer:

(a) 17634.24 Ω

(b) 0.0068 A

Explanation:

(a)

The formula for inductive inductance is given as

X' = 2πFL................... Equation 1

Where X' = inductive reactance, F = frequency, L = inductance

Given: F = 60 Hz, L = 46.8 H, π = 3.14

Substitute into equation 1

X' = 2(3.14)(60)(46.8)

X' = 17634.24 Ω

(b)

From Ohm's law,

Vrms = X'Irms

Where Vrms = Rms Voltage, Irms = rms Current.

make Irms the subject of the equation

Irms = Vrms/X'...................... Equation 2

Given: Vrms = 120 V, X' = 17634.24 Ω

Substitute into equation 2

Irms = 120/17634.24

Irms = 0.0068 A

Answer:

B) A person on the moving Earth observing the light from a distant star

Explanation:

The aberration of starlight is a phenomenon in Astronomy used to describe the 'seeming' or apparent movement of a star around its true position. This seeming' or apparent movement of a star is caused by and dependent on the velocity of the observer. The movement of the observer relative to the star creates the notion that the star is moving. However, in reality, the star is static.  The star is not moving, what is really moving is the observer; this movement of the observer is what makes it seem that the star was moving. It is not an optical illusion, although, the effect seems close enough.

For example, in the analogy given:

A car moving in the rain or under the rainfall

The car is the one moving in the same way that the Earth is the one revolving. The rainfall drops vertically in the same way that the starlight is static. The movement of the car relative to the rainfall is what makes it seem that the rain is falling 'diagonally'. The same visual analogy is observe when a person on the moving Earth observing the light from a distant star

Hence, the correct option is B

Answer:

Correct answers:

Explanation:

• It is a satellite that collects data about rain and snow.

GPM is part of NASA's Earth Systematic Missions program and its function is to track precipitation from space with satellites and to provide accurate information of the time, place an amount of rain or snow anywhere in the world.

• Its orbit covers 90 percent of Earth's surface.

 GPM has unique perspective on measuring precipitation from space because of their collected satellite retrievals.

• The sensors measure microwaves.

GPM Microwave Imager (GMI) is a sensor that observes the microwave energy emitted by the Earth and atmosphere, so it improves the accuracy of rain and snowfall estimates.

Answer:

3.6 N

Explanation:

E=F/q

so

F=qxE

F= 9x10^-5x4x10^4

F= 3.6 N

The increase in potential energy of the spring when compressed a distance of 0.062m from its unstretched position, with a spring constant of 87600N/m, is approximately 0.168 kJ.

The increase in potential energy of the spring can be calculated using the formula U = ½kx². In this problem, k is the spring constant which is 87,600 N/m and x is the displacement of the leaf spring when it is compressed, which is 0.062 m. Plugging in these values, we get U = ½ * (87,600 N/m) * (0.062 m)² = 167.5424 Joules. This value is in Joules, so to convert it to kilojoules (kJ), we divide by 1000. So, the increase in potential energy of the spring is approximately 0.168 kJ.

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Answer:

2.9 m/s

Explanation:

The speed of the two-block system center of mass will not be changed by the collision, since it is an internal force. The only force acting upon the two-block system, then, will be gravity. We can calculate the effects of gravity upon the speed of the blocks along the slope by taking the sine of the angle of the slope (37°) and multiplying it by the magnitude of Fg in the vertical direction. sin(37°)*(3.92 N) = 2.36 N. This force acts upon the system and accelerates it down the slope, which can be modeled with the equation 2.36 N = (0.40 kg)*a. Solving for a, we find that gravity accelerates the block at 5.9 m/s2 along the slope of the block. Using this figure, we can find the speed of the system with the equation v = v0 + at. Initial velocity of the system is zero, as both blocks are moving towards each other at equal speed, so v = (5.9 m/s2)*(0.50 s). Velocity of the system after 0.50 seconds = 2.9 m/s.

The center-of-mass of an isolated two-block system moving towards each other at the same speed remains the same before the collision; therefore, the center-of-mass velocity is also 0.35 m/s just before the blocks collide.

In this scenario, we are dealing with the principle of conservation of momentum in an isolated system. In two-dimensional collisions involving identical masses moving toward each other at the same speed, the center-of-mass velocity does not change due to the collision.

To find the speed of the two-block system's center of mass, we can use the formula for center-of-mass velocity, that is the weighted average of the individual velocities. Since the blocks have identical mass, the center-of-mass velocity is just the average of their individual velocities. Given both blocks are moving toward each other at a speed of 0.35 m/s, the speed of the center-of-mass of the system before the blocks collide is also 0.35 m/s.

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Answer:

33.1 degrees celcius

Explanation:

Look up attached file

The equilibrium temperature of the train’s top cannot be calculated without the specific heat transfer due to convection. The Stefan-Boltzmann law allows us to calculate heat transfer from radiation but not convection.

The key to solving this problem is the Stefan-Boltzmann law of radiation, which describes the rate of heat transfer by emitted radiation. In brief, P = σeAT⁴, where σ = 5.67 × 10⁻⁸ J/s.m².K⁴ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, and T is its temperature in Kelvins.

Given that the top surface of the car absorbs solar radiation at a rate of 200 W/m² and its dimensions, we can calculate the area of the surface. The surface, if deemed perfectly insulated, only loses heat through convection, not radiation. However, the amount of heat transfer due to convection is not provided in the question, making it impossible to calculate the exact equilibrium temperature of the top surface of the car.

The net rate of heat transfer takes into account both the temperature of the object and the temperature of its surroundings. If T₁ represents the temperature of the surface to be investigated, and T₂ the ambient temperature, without the value of heat lost through convection, we cannot calculate T₁. Therefore, more information would be required to answer your question fully.

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Final answer:

The direction of the resultant force acting on an electron moving upward on the page through a magnetic field directed into the page is to the left, as per the left-hand rule.

Explanation:

The question deals with the Lorentz force experienced by a charged particle, in this case an electron, moving through a magnetic field. When an electron with initial velocity directed from the bottom edge to the top edge of the page moves through a uniform magnetic field directed into the page, the force acting on it can be determined using the left-hand rule, since electrons have a negative charge. Holding your left thumb up (direction of the electron's velocity) and the fingers into the page (direction of the magnetic field), the resultant force will be directed to the left of the page. Therefore, the correct answer to the direction of the resultant force acting on the electron is c. to the left.

Answer:

The change in momentum for average force is: Δ[tex]P_{average force} = F_{avg}[/tex]Δt and it is calculated below.

For thick and thin spring, the %difference between change in momentum for average force and change in momentum for velocities is; 178.27% and 159.72% respectively

Explanation:

The average force applied to an object can be calculated by multiplying the average net external force by the time interval over which the force acts. The change in momentum can be calculated using the average force, as well as the change in velocity. The percentage difference between the two values can be found by dividing the difference by the change in momentum using the change in velocity and then multiplying by 100.

The average force applied to an object can be calculated using the average net external force multiplied by the time interval over which the force acts. This relationship is known as the impulse-momentum theorem, which states that the change in momentum is equal to the impulse. Impulse is the product of the average force and the time interval.

To calculate the change in momentum of an object using the average force, you can multiply the average force by the duration the force is applied. This will give you the change in momentum caused by the force. You can then calculate the change in momentum using the change in velocity by multiplying the mass of the object by the change in velocity. Finally, you can find the percentage difference between the two values by dividing the difference between the two values by the change in momentum using the change in velocity and then multiplying by 100.

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Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

Answer:

a)9.8

Explanation:

because any object falling from any height is 9.8

(sorry that's the only one i know)

1. The speed of the rock relative to the car when it hits the windshield is [tex]\( {42.00 \, \text{m/s}} \).[/tex]

2. The distance above the point of contact with the windshield from which the rock was dropped is [tex]\( {67.5 \, \text{m}} \).[/tex]

3. The driver would have had [tex]\( {3.71 \, \text{s}} \)[/tex] to react before the rock strikes the windshield.

Let's solve each part of the problem step-by-step.

Part (a): Find the speed of the rock relative to the car when it hits the windshield.

Given:

- Car speed, [tex]\( v_{\text{car}} = 21 \, \text{m/s} \)[/tex]

- Windshield angle with the horizontal,[tex]\( \theta = 60^\circ \)[/tex]

Since the rock is dropped from rest, its initial vertical velocity is [tex]\( 0 \, \text{m/s} \).[/tex]

The rock falls under the influence of gravity, so its vertical velocity [tex]\( v_{\text{vertical}} \)[/tex] at any time t is given by:

[tex]\[ v_{\text{vertical}} = g t \][/tex]

where [tex]\( g = 9.8 \, \text{m/s}^2 \).[/tex]

The horizontal velocity of the rock is [tex]\( 0 \, \text{m/s} \)[/tex] with respect to the ground.

To find the speed of the rock relative to the car when it hits the windshield, we need to consider the components of velocity with respect to the car:

1. The car's horizontal velocity component.

2. The rock's vertical velocity component.

The velocity components with respect to the car:

- Horizontal component relative to the car, [tex]\( v_{\text{horizontal, relative}} = v_{\text{car}} \)[/tex]

- Vertical component, [tex]\( v_{\text{vertical}} = g t \)[/tex]

The speed of the rock relative to the car can be found using the Pythagorean theorem:

[tex]\[ v_{\text{relative}} = \sqrt{(v_{\text{horizontal, relative}})^2 + (v_{\text{vertical}})^2} \][/tex]

We need to find t when the rock strikes the windshield.

Since the rock strikes the windshield at a right angle, and the windshield makes a 60° angle with the horizontal, the relative speed of the rock [tex]\( v_{\text{relative}} \)[/tex]  must have components that align with the 60° angle.

Thus:

[tex]\[ v_{\text{relative, vertical}} = v_{\text{relative}} \sin(60^\circ) \][/tex]

[tex]\[ v_{\text{relative, horizontal}} = v_{\text{relative}} \cos(60^\circ) \][/tex]

Given:

[tex]\[ v_{\text{relative, vertical}} = v_{\text{vertical}} = g t \][/tex]

[tex]\[ v_{\text{relative, horizontal}} = v_{\text{car}} = 21 \, \text{m/s} \][/tex]

From the component relationship:

[tex]\[ \frac{v_{\text{relative, vertical}}}{v_{\text{relative, horizontal}}} = \tan(60^\circ) = \sqrt{3} \][/tex]

[tex]\[ \frac{g t}{21} = \sqrt{3} \][/tex]

[tex]\[ g t = 21 \sqrt{3} \][/tex]

[tex]\[ t = \frac{21 \sqrt{3}}{9.8} \][/tex]

Calculate t:

[tex]\[ t \approx \frac{21 \cdot 1.732}{9.8} \approx \frac{36.372}{9.8} \approx 3.71 \, \text{s} \][/tex]

Now, we can find the vertical velocity [tex]\( v_{\text{vertical}} \)[/tex]:

[tex]\[ v_{\text{vertical}} = g t = 9.8 \times 3.71 \approx 36.36 \, \text{m/s} \][/tex]

Now, the speed of the rock relative to the car:

[tex]\[ v_{\text{relative}} = \sqrt{(21)^2 + (36.36)^2} \approx \sqrt{441 + 1322.49} \approx \sqrt{1763.49} \approx 42.00 \, \text{m/s} \][/tex]

Part (b): Find the distance above the point of contact with the windshield from which the rock was dropped.

The distance h from which the rock was dropped can be found using the kinematic equation:

[tex]\[ h = \frac{1}{2} g t^2 \][/tex]

Using [tex]\( t \approx 3.71 \, \text{s} \)[/tex]:

[tex]\[ h = \frac{1}{2} \times 9.8 \times (3.71)^2 \approx \frac{1}{2} \times 9.8 \times 13.76 \approx 67.5 \, \text{m} \][/tex]

Part (c): The time available for the driver to react is the time \( t \) it takes for the rock to fall, which we calculated in part (a):

[tex]\[ t \approx 3.71 \, \text{s} \][/tex]

Answer:

Explanation:

General guidance

Concepts and reason

The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.

Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.

Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.

Fundamentals

The condition for diffraction maximum in the case of single slit diffraction is as follows:

sin Θ=λ/α

Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .

The incorrect options are as follows:

• The central bright fringe remains in same size.

The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.

• The effect cannot be determined unless the distance between the slit and screen is known.

Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.

• The central bright fringe becomes narrower.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.

If the width of the slit increases, then the central bright fringe width also increases.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.

The condition for diffraction maximum is as follows:

sin  Θ=λ/α

The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

Therefore,

• The central bright fringe becomes wider is correct.

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.  

Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.

If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.

Answer

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.

Answer:

1) μ = 1.33 10⁻³ kg / m , F = - 14,256 ,  2) v= 103.53 m/s, 3)  f = 138.04 Hz , 4)  1, 25, 50, 76, 101   , 5) A = 0.00869 m , 6)  # _position = (# _account-1) (1.5m / 100 accounts)

Explanation:

1) Linear density is the mass per unit length

     μ = m / L

     μ = 2 1 10⁻³ / 1,5

     μ = 1.33 10⁻³ kg / m

this is the density when the chain is stretched, which is when the pulse occurs

we can find the tension with

     F = - k (x₁-x₀)

where k is the spring constant

     F = - 28.8 (1.5 -1.005)

     F = - 14.256 N

the negative sign indicates that the force is restorative

2) the pulse speed is

      v = √ T /μ

      v = √ 14,256 / 1,33 10⁻³

      v = 103.53 m / s

3) If standing waves are formed with fixed points at the ends and 4 antinodes, the wavelength is

          2 λ = L

            λ = L / 2

wave speed is related to frequency and wavelength

           v = λ f

            f = v / λ

            f = v 2 / L

            f = 103.53 2 / 1.5

            f = 138.04 Hz

4) The marbles are numbered, the marbles that remain motionless are

   the first (1) and the last (101)

Let's look for the distance to each node, for this we must observe that in each wavelength there is a node at the beginning, one in the center and one at the end, therefore the nodes are in

         #_node = m λ / 2 = m L / 4

        #_node     position (m)

         1                  1.5 / 4 = 0.375

         2             2 1.5 / 4 = 0.75

         3             3 1.5 / 4 = 1,125

Since there are 101 marbles in the initial length, this number does not change with increasing length, so there is 101 marble in 1.5 m. Let's find with a direct proportion rule the number of marbles at these points with nodes

        #_canica = 0.375 m (101 marble / 1.5 m) 0.375 67.33

        # _canica = 25

        #_canica = 0.75 67.33

        #_canica = 50

        # _canica = 1,125 67.33

        #canica = 75.7 = 76

in short the number of the fixed marbles is

      1, 25, 50, 76, 101 canic

5) The movement of the account is oscillatory at this point, which is why it is described by

          y = A cos wt

          [tex]v_{y}[/tex]= -A w sin wt

the speed is maximum for when the breast is worth ±1

          v_{y} = Aw

           A = v_{y} / w

angular velocity related to frequency

         w = 2π f

          A = v_{y} / 2πf

          A = 7.54 / (2π 138.04)

          A = 0.00869 m

6) for the position of each account we can use a direct proportion rule

      in total there are 100 accounts distributed in the 1.50 m distance, the #_account is in the # _position. Note that it starts to be numbered 1, so this number must be subtracted from the index of the amount

       # _position = (# _account-1) (1.5m / 100 accounts)

#_canic position(m)

   1          0

   2         0.015

   3         0.045

   4         0.06

7) the wave has a constant velocity, but every wave is oscillated perpendicular to this velocity, with an oscillatory movement described by the expression

         y = Acos wt

the maximum speed is

         [tex]v_{y}[/tex] = -Aw sin wt

speed is maximum when the sine is ±1

         v_{y} = A w

to calculate the amplitude of the count we use that for a standing wave

         y = 2Asin kx

          y / A = 2 sin (2π /λ x)

the wavelength is

 λ = 0.75 m

the position is

x (30) = 29 1.5 / 100 = 0.435  m

          y (30) A = 2 sin (2pi 0.435 / 0.75)

          y (30) / A = 0.96 m

Answer:

Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m

Explanation:

From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month

Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds

Now, the square of the period of any planet is given by the formula;

T² = (4π²/GM)α³

Where;

G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²

M is mass of sun which has a value of 1.99 x 10^(30) kg

α is the semi major axis of the comets orbit.

Thus, making α the subject, we have;

α = ∛(GMT²/4π²)

So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)

α = 1.89 x 10^(13) m

Answer:

Explanation:

the picture attached shows the whole explanation

The Earth is slightly flatter at the poles, making the correct answer option C. It is an oblate spheroid with a denser core made primarily of iron and nickel.

The correct answer to the question, 'The earth is:', is C: slightly flatter at the poles. This is because the Earth's shape is an oblate spheroid, meaning it's somewhat squished at the poles and bulges a bit at the equator due to its rotation. The Earth's equatorial radius is greater than its polar radius, which results in this slight flattening at the poles. Also, the Earth's core is composed predominantly of iron and nickel, making it very dense.

While it is true that the Earth is not a perfect sphere, it is also not thinner at the equator as option B suggests. Option D mentions that Earth is made of mostly carbon and iron, which is incorrect; the Earth is composed primarily of iron, oxygen, silicon, and other elements, but not mostly carbon.

Answer:

a) = 6.23eV

b) = 0eV

c) = 6.23V

d) = 304.12nm

Explanation:

λ = 199nm

E = 4.08eV = Φ

hf = K(max) + Φ

Φ = work function of the target material

hf = photon energy

K(max) = k.e of the metal

K(max) = [(6.626*10⁻³⁴) * (3.0*10⁸)] / 199*10⁻⁹

K(max) = 9.99*10⁻¹⁹J

1.602*10⁻¹⁹J = 1eV

9.99*10⁻¹⁹ = 6.23eV

K(max) = 6.23eV

b). The slowest moving electron has a kinetic energy of zero

c). The stopping potential is the potential difference required to stop the fastest electron.

eV = K(max)

V = K(max) / e

V = 6.23eV / e

V = 6.23V

d)

The cut- off wavelength occurs when the maximum K.E is zero.

Φ = hc / λ

λ₀ = hc / Φ

λ₀ = (6.626*10⁻³⁴ * 3.0*10⁸) / (4.08 * 1.602*10⁻¹⁹)

λ₀ = 1.9878*10⁻²⁵ / 6.5361*10⁻¹⁹

λ₀ = 3.04*10⁻⁷m

λ₀ = 304.12nm

Answer:

Approximately 4.5 square miles

Explanation:

(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)

Final answer:

The 'orange line' referred to in the question likely represents data in a graph related to either the wavelength or frequency of orange light in the visible spectrum or a trend in statistical analysis. The exact answer depends on the specific context of the graph or data set from which the 'orange line' is derived.

Explanation:

The question 'The orange line is measuring the' seems to be related to analyzing data, most likely from a graph or chart that measures certain variables, potentially in the context of physics or mathematics dealing with light spectra or statistical analysis. Given the details provided, including mention of the visible light spectrum and specific wavelengths denoted by numbers, it appears that the 'orange line' may refer to a line on a graph that represents the wavelength or frequency of the color orange within the visible light spectrum.

However, the context provided is ambiguous. For instance, the chapter headings and the Conclusion provided suggests statistical analysis, perhaps in the field of queueing theory and how waiting times are affected by different systems, which implies a more mathematical or business-oriented question. In the field of mathematics, 'orange line' could also be a reference to a trend line or data line in a statistical plot.

Answer:

Their final velocity is 0.122m/s

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision. The bodies move with a common velocity after collision.

Momentum = mass × velocity

BEFORE COLLISION

momentum of the first train

= 1.5×10^5 × 0.3

= 0.45×10^5kgm/s

Momentum of second train

= 1.1×10^5 × (-0.12) (velocity is negative since the body is moving towards the left)

= -0.132 × 10^5kgm/s

AFTER COLLISION

Momentum of both bodies is given as:

(1.5×10^5+1.1×10^5)V

= 2.6×10^5V

V is their common velocity

According to the law:

0.45×10^5+(-0.132×10^5) =2.6×10^5V

0.45×10^5 - 0.132×10^5 = 2.6×10^5V

0.318×10^5 = 2.6×10^5V

0.318 = 2.6V

V = 0.318/2.6

V = 0.122m/s

Answer:

A) speed = 3144287.425 m/s

B) period = 8.57x10^-8 s

C) KE = 3.302x10^-14 J

D) potential difference = 103187.5 V

Explanation:

Detailed explanation and calculation is shown in the image below.

Answer:

Horizontal launch

[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]

Vertical launch

[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]

Explanation:

The launch speed is calculated by means of the Principle of Energy Conservation:

[tex]U_{k} = K[/tex]

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]v = x \cdot \sqrt{\frac{k}{m} }[/tex]

[tex]v = (0.15\,m)\cdot \sqrt{\frac{100\,\frac{N}{m} }{0.3\,kg} }[/tex]

[tex]v \approx 2.739\,\frac{m}{s}[/tex]

The velocities for each scenario are presented herein:

Horizontal launch

[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]

Vertical launch

[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]

Final answer:

The launch velocity of the T-shirt, when the spring with a spring constant of 100 N/m is compressed by 0.15 m, is calculated using conservation of energy. It turns out to be approximately 2.74 m/s for both the horizontal and vertical launches.

Explanation:

To calculate the launch velocity of the T-shirt from the cannon in both horizontal and vertical scenarios, we'll use the conservation of energy principle. This principle states that the potential energy stored in the compressed spring is converted into kinetic energy of the moving T-shirt.

Calculation of Launch Velocity (Horizontal and Vertical)

The energy stored in the compressed spring can be given by the formula:
[tex]E = 1/2 k x^2[/tex]
where E is the energy in joules, k is the spring constant, and x is the compression distance.

For the given spring with k = 100 N/m compressed by 0.15 m, the energy stored is:
[tex]E = 1/2 (100) (0.15)^2 = 1.125 joules[/tex]

Now we set this equal to the kinetic energy of the T-shirt upon release:
[tex]KE = 1/2 m v^2[/tex]
where KE is the kinetic energy, m is the mass of the T-shirt, and v is the velocity.

To find the launch velocity, we solve for v given the mass m = 0.3 kg and energy E = KE:

[tex]v = sqrt((2 * E) / m) = sqrt((2 * 1.125) / 0.3) = sqrt(7.5) = approximately 2.74 m/s[/tex]

This velocity will be the same for both the horizontal and the vertical launch as friction is ignored and the energy is purely transferred into the kinetic energy of the T-shirt.

Answer:

Explanation:

Given that,

Number of turns

N = 43 turns

Dimensions of rectangle

Length L = 10cm = 0.1m

Width W = 5cm = 0.05

Then, area of rectangle

A = L×W = 0.1 × 0.05 = 0.005m²

Initial magnetic field

Bi = 0T

Final magnetic field

Bf = 0.5T

The displacement time

∆t = 0.431s

Induce emf ε?

EMF is given as

ε = —NdΦ/dt

Where magnetic flux

Φ = BA

Note A is constant

Then,

ε = —NdΦ/dt = —Nd(BA)/dt

ε = —N•A•dB/dt

ε = —N•A•(∆B/∆t)

ε = —N•A•(Bf - Bi) / (∆t)

ε = —43 × 0.005 × (0.5 - 0) /0.431

ε = —43 × 0.005 × 0.5 /0.431

ε = —0.249 V

The average induced EMF is 0.249V

Answer:

A.True

B. True

C False

D. False

E. True

Explanation:

The potential at x = 1.0 m in the given uniform electric field, with the given parameters, would be 400 V.

The subject question pertains to the concept of electric potential in a uniform electric field. The potential difference, V, in an uniform electric field is given by the equation V = Ed, where E is the field strength and d is the distance. Given an electric field strength, E, of 650 N/C and a potential, V, at x = 3.0 m of 1 700 V, we seek the potential at x = 1.0m. Since the electric field is uniform, we know that the change in potential is linear with distance. Because the distance decreases by 2 meters (from x = 3.0 m to x = 1.0 m), the potential will decrease by E*2, which is 650 N/C * 2 m = 1300 V. So, the potential at x = 1.0 m is 1 700 V - 1 300 V = 400 V.

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To find the electric potential at x = 1.0 m in a uniform electric field directed along the x-axis, subtract the potential difference due to the field over the distance moved from the initial potential; the answer is 400 V.

The question is asking to calculate the electric potential at a certain point given a uniform electric field and an initial electric potential at a different point. To find the potential at x = 1.0 m, we can use the relationship between electric field (E), electric potential (V), and distance (d), with the following formula: V = V_0 - E * d, where V is the potential at the point of interest, V_0 is the initial potential, E is the electric field strength, and d is the distance between the points.

Gor the given electric field of 650 N/C directed parallel to the positive x-axis, and with the potential at x = 3.0 m being 1700 V, the potential difference \\Delta V caused by moving from x = 3.0 m to x = 1.0 m can be calculated by multiplying the electric field strength by the distance over which the field is applied, which is 2.0 m (3.0 m - 1.0 m). Hence, \\Delta V = E * d = 650 N/C * 2.0 m = 1300 V.

Therefore, the potential at x = 1.0 m is V = 1700 V - 1300 V = 400 V.

Answer:

10

Explanation:

v=f×lander

20=f2

f=10

Answer:

31,360J

Explanation:

Gravitation potential energy (gpe) is calculated from the formula mgh.

That implies, gpe = mgh

Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J

Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.

The gravitational potential energy of the skier at the top of the slope, calculated using the formula PE = mgh, is 31,360 joules.

To find the gravitational potential energy of the skier, we use the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration because of gravity, and h is the height. The acceleration because gravity is 9.8 m/s².

So, we substitute these values into the formula:

PE = (80 kg) × (9.8 m/s²) × (40 m) = 31,360 joules.

Therefore, the gravitational potential energy of the skier before she skips down the slope is 31,360 joules.

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Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

                           E = Vo / D

                           E = 600 / 0.3

                           E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

                           Fe = E*q

                           Fe = (2,000 V / m) * ( 3*10^-5 C)

                           Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

                          Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

                          Fe = m*a

                          a = Fe / m

                          a = 0.06 / 0.0004

                          a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

                         s = C - A

                         s = ( 0.25 , 12 ) - ( 0.05 , 12 )

                         s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

                        vf^2 = vi^2 + 2*a*s

                        vf^2 = 0 + 2*0.2*150

                        vf = √60

                        vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

                       vf - vi = a*t

                       t = ( 7.746 - 0 ) / 150

                       t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

                      [tex]Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)][/tex]

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

                     [tex]E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\[/tex]

- Again, apply the Newton's second law of motion and determine the acceleration (a):

                     Fe = m*a

                     a = Fe / m

                     [tex]a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}[/tex]

- Where the acceleration is rate of change of velocity "dv/dt":

                     [tex]\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\[/tex]

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

                      [tex]c = \frac{A*eo}{d}[/tex]

Where, eo = 8.854 * 10^-12  .... permittivity of free space.

                     [tex]K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\[/tex]

- The differential equation turns out ot be:

                     [tex]\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\[/tex]

- Separate the variables the integrate over the interval :

                    t : ( 0 , t )

                    v : ( 0 , vf )

Therefore,

                   [tex]\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )[/tex]

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

                    [tex]vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s[/tex]

- The final velocity of particle while charging the capacitor would be:

                   vf = 7.264 m/s ... slightly less for the fully charged capacitor