You sit at the middle of a large turntable at an amusement park as it begins to spin on nearly frictionless bearings, and then spins freely. When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged, and why

Answer:

B. 0.224 m/s²

Explanation:

Given:

Mass of the object (m) = 20 kg

The forces acting on the object are:

[tex]\vec{F_1}=3\vec{i}\ N\\\\\vec{F_2}=5\vec{j}\ N\\\\\vec{F_3}=(\vec{i}-3\vec{j})\ N[/tex]

Now, the net force acting on the object is equal to the vector sum of the forces acting on it. Therefore,

[tex]\vec{F_{net}}=\vec{F_1}+\vec{F_2}+\vec{F_3}\\\\\vec{F_{net}}=3\vec{i}+5\vec{j}+\vec{i}-3\vec{j}\\\\\vec{F_{net}}=(3+1)\vec{i}+(5-3)\vec{j}\\\\\vec{F_{net}}=(4\vec{i}+2\vec{j})\ N[/tex]

Now, the magnitude of the net force is equal to the square root of the sum of the squares of its components and is given as:

[tex]|\vec{F_{net}}|=\sqrt{4^2+2^2}\\\\|\vec{F_{net}}|=\sqrt{20}\ N[/tex]

Now, from Newton's second law, the magnitude of acceleration is equal to the ratio of the magnitude of net force and mass. So,

Magnitude of acceleration is given as:

[tex]|\vec{a}|=\dfrac{|\vec{F_{net}}|}{m}\\\\|\vec{a}|=\frac{\sqrt{20}\ N}{20\ kg}\\\\|\vec{a}|=0.224\ m/s^2[/tex]

Therefore, option (B) is correct.

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

The work done pumping the liquid out of the tank relies on the weight of the mass being lifted, the gravity, and the height it is being lifted to. Calculations are given for lifting all the fluid and two-thirds of it to two different heights.

To calculate the work done pumping the liquid out of the tank, we first need to find the volume of the tank, which is 4 meters * 3 meters * 6 meters, giving a total of 72 cubic meters of rubbing alcohol. Multiplying this by the density of the alcohol (786 kg/m^3) gives us the total mass of the alcohol, 56,592 kg. The work done by gravity when an object is lifted is equal to the weight of the object (mass*gravity) multiplied by the distance it is lifted (height). Therefore, we can calculate the work done pumping the liquid out of the tank:

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The work done to pump all of the liquid out over the top of the tank is calculated as 1665568.8 J, and 4441516.8 J through a spout 2 meters above the tank. To pump out two-thirds of the liquid over the top, the work is 1109203.2 J, and through the spout, it is 2957875.2 J. Each part uses the weight of the liquid and the distance it needs to be moved to calculate the work done.

Part a :

To find the work done, we need to calculate the force required to move the liquid to the top of the tank and the distance it needs to be moved.

Calculate the volume of the tank:

Volume = length × width × depth = 4 m × 3 m × 6 m = 72 m³.

Calculate the mass of the liquid:

Mass = density × volume = 786 kg/m³ × 72 m³ = 56652 kg.

Calculate the weight of the liquid:

Weight = mass × gravity = 56652 kg × 9.8 m/s² = 555189.6 N.

Calculate the center of mass:

The center of mass for a full tank is at half the depth:

3 meters.

Calculate the work done:

Work = weight × height = 555189.6 N × 3 m = 1665568.8 J (Joules).

Part b :

Here, the height the liquid needs to be moved is 6 m (tank height) + 2 m (spout height):

6 + 2 = 8 meters.

Work = weight × height

Work = 555189.6 N × 8 m = 4441516.8 J.

Part c :

Calculate the volume of two-thirds of the tank:

Volume = (2/3) × 72 m³ = 48 m³.

Calculate the mass of two-thirds of the liquid:

Mass = 786 kg/m³ × 48 m³ = 37728 kg.

Calculate the weight of two-thirds of the liquid:

Weight = 37728 kg × 9.8 m/s² = 369734.4 N.

Since the tank is still of uniform depth, the center of mass for the remaining liquid will be halfway up the tank's current depth:

1.5 meters (half of the 3 meters left).

Work = weight × height

Work = 369734.4 N × 3 m = 1109203.2 J.

Part d :

Height the liquid needs to be moved = 6 m (tank height) + 2 m (spout height) = 8 meters.

Work = weight × height

Work = 369734.4 N × 8 m = 2957875.2 J.

Answer:

Complete question

In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.095 T magnetic field.

a. What electric field strength, in volts per mater, is needed to select a speed of 4.2 x 10^6 m/s?

b. What is the voltage, in kilovolts, between the plates if they are separated by 0.95 cm?

Explanation:

Given that,

magnetic field B = 0.095T

Speed of particle v = 4.2 ×10^6m/s

Separation between plate d = 0.95cm

d = 0.95/100 = 0.0095m

a. Using the mass spectrometer velocity selector relationship between the electric field and magnetic field.

v = E/B

Where

v is the speed selector

B is magnetic field

E is electric field

Therefore, E = vB

E = 4.2 × 10^6 × 0.095

E = 0.399× 10^6

E = 3.99 × 10^5 V/m

b. Voltage?

The relationship between electric field and potential difference between the two plates is given as

V = Ed

V = 3.99 × 10^5 × 0.0095

V = 3790.5 V

To kV, 1kV = 1000V

Then, V = 3.7905kV

V ≈ 3.791 kV

The electric field strength needed in a mass spectrometer to select a velocity of 4.00 × 106 m/s with a 0.100-T magnetic field is 400,000 N/C. The voltage required across plates separated by 1.00 cm is 4,000 V.

In a velocity selector within a mass spectrometer, a charged particle remains undeflected when the electric force equals the magnetic force. This condition is given by qE = qvB, where q is the charge, E is the electric field strength, v is the velocity of the particle, and B is the magnetic field strength. The question asks for the electric field strength needed to select particles with a velocity of 4.00 × 106 m/s when subjected to a 0.100-T magnetic field. Using the condition for balance, E = vB, we can substitute the given values to find E = 4.00 × 106 m/s × 0.100 T = 400,000 N/C. For part (b), voltage (V) can be determined by the relation V = Ed, where d is the separation between the plates. For a plate separation of 1.00 cm, or 0.01 m, the voltage is V = 400,000 N/C × 0.01 m = 4,000 V.

Final answer:

The acceleration of the child in the wagon can be calculated using Newton's Second Law of Motion. The sum of the forces acting on the system is calculated by adding the forces exerted by the children and subtracting the force of friction. Using the given values, the acceleration can be determined by dividing the sum of the forces by the mass of the system. If the force of friction is increased to 20.0 N, the acceleration can be recalculated using the new value.

Explanation:

To calculate the acceleration of the child in the wagon, we can use Newton's Second Law of Motion:

ΣF = ma

Where ΣF is the sum of all the forces acting on the system, m is the mass of the system, and a is the acceleration.

In this case, the forces acting on the system are the forces exerted by the two children and the force of friction.

Using the given values:

Force1 = 75.0 N

Force2 = 95.0 N

Friction = 12.0 N

Mass = 21.0 kg

The sum of the forces acting on the system is:

ΣF = Force1 + Force2 - Friction

Substituting the values:

ΣF = 75.0 N + 95.0 N - 12.0 N

ΣF = 158.0 N

Now we can calculate the acceleration:

a = ΣF / m

Substituting the values:

a = 158.0 N / 21.0 kg

a = 7.52 m/s²

Therefore, the acceleration of the child in the wagon is 7.52 m/s².

(d) If friction is 20.0 N, we can recalculate the acceleration using the new value:

ΣF = 75.0 N + 95.0 N - 20.0 N

ΣF = 150.0 N

a = ΣF / m

a = 150.0 N / 21.0 kg

a = 7.14 m/s²

Therefore, if friction is 20.0 N, the acceleration of the child in the wagon would be 7.14 m/s².

Answer:

0.26086 Hz

Explanation:

solution:

The relative speed of the wave is:

v = 4.9+1.1 = 6 m/s

the frequency is:

f = v/λ

 = 6/23

 = 0.26086 Hz

Answer:

(a) 21.44 ft/s

(b) 0 ft/s

(c) 19.51 ft/s

Explanation:

2 in = 2/12 ft = 0.167 ft

For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:

[tex]v(r) = v_c\left[1 - \frac{r^2}{R^2}\right][/tex]

where R = 0.167 ft is the pipe radius and [tex]v_c[/tex] is the constant fluid velocity at the center of the pipe.

We can integrate this over the cross-section area of the in order to find the volume flow

[tex]\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\[/tex]

So the average velocity

[tex]v = \dot{V} / A = v_c/2 = 10.72[/tex]

[tex]v_c = 10.72*2 = 21.44 ft/s[/tex]

b) At the wall of the pipe, r = R so [tex]v(R) = v_c(1 - 1) = 0 ft/s[/tex]

c) At a distance of 0.6 in = 0.6/12 = 0.05 ft

[tex]v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s[/tex]

Answer:

The answers to the questions are;

(a) The velocity at the center of the pipe is 21.44 ft/s

(b) The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is 19.63 ft/s.

Explanation:

To solve the question, we note that

The velocity profile in the cross section of a circular pipe with laminar flow is given by

U = 2×v×[1 - (r/r₀)²]

Where

U = The sought velocity at a point

r = Pipe radius where velocity is sought

r₀ = Internal radius of pipe = for  2-in Schedule 40 steel pipe =  2.067 in 52.6 mm

v = Average velocity of flow = 10.72 ft/s = 3.2675 m/s

Therefore we have

(a) The velocity at the center of the pipe

At the center r = 0 so we have

U = 2×v×[1 - (r/r₀)²]

At center U = 2×10.72 ft/s×[1 - (0/2.067 in)²] =  2×10.72 ft/s = 21.44 ft/s

(b) The velocity at the wall of the pipe is given by

r = r₀ ⇒ U = 2×v×[1 - (r/r₀)²] ⇒ U = 2×v×[1 - (r₀/r₀)²]

= U = 2×v×[1 - (1)²] =  2×v×0 = 0

The velocity at the wall of the pipe is 0 ft/s

(c) The velocity at a distance of 0.6 inches from the center-line is given by

U = 2×v×[1 - (r/r₀)²] = 2×10.72 ft/s×[1 - (0.6/2.067)²]  =  19.63 ft/s.

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

The induced electromotive force (EMF) in a generator coil is directly proportional to the area of the coil, frequency of rotation, and the magnetic field strength, but not to the coil's internal resistance. Changing any of these factors as described, except resistance, results in a proportional change in the induced EMF.

The behavior of induced electromotive force (EMF) in a generator is explained by Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is proportional to the rate of change of magnetic flux through the coil.

These principles help us to understand the operation of an electric generator, where mechanical work is converted into electrical energy.

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Approximately 1.6 million joules of work is required to pump the liquid out of the spherical tank through a 1 m long spout at the top.

The work done to pump the liquid out of a tank can be calculated using the formula W = ρgVh, where ρ is the density of the liquid, g is the acceleration due to gravity, V is the volume of the liquid, and h is the height up to which the liquid is pumped. Since the tank is spherical and half full, the volume of the liquid is ½(4/3πr³), or 2πr³. Substituting the given values: ρ = 900 kg/m³, g = 9.8 m/s², r = 3 m, and h is the radius of the sphere plus the length of the spout (3 m + 1 m = 4 m), we get W = 900 kg/m³ * 9.8 m/s² * 2π(3 m)³ * 4 m ≈ 1.6 * 10⁶ J. Therefore, approximately 1.6 million joules of work is required to pump the liquid out of the tank.

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Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=[tex]v=6.1\times 10^7m/s[/tex]

Charge on electron, [tex]q=e=1.6\times 10^{-19} C[/tex]

Mass of electron,[tex]m_e=9.1\times 10^{-31} kg[/tex]

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

[tex]V=\frac{v^2m_e}{2e}[/tex]

Where V=Potential difference

[tex]m_e=[/tex]Mass of electron

v=Velocity of electron

Using the formula

[tex]V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}[/tex]

[tex]V=10581.59 V[/tex]

Hence, the potential difference=10581.59 V

Final answer:

To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, the potential difference required in the first part of the experiment is approximately 88.6 volts.

Explanation:

To accelerate electrons to a speed of 6.1 × 10^7 m/s in a constant magnetic field with a strength of 0.65 T, we need to calculate the potential difference required in the first part of the experiment. The formula for the potential difference is given by:

V = (1/2)m*(v^2)/(q * B)

Where V is the potential difference, m is the mass of the electron (9.11 × 10^-31 kg), v is the velocity of the electron (6.1 × 10^7 m/s), q is the charge of the electron (-1.6 × 10^-19 C), and B is the magnetic field strength (0.65 T).

Plugging in the values into the formula, we get:

V = (1/2)(9.11 × 10^-31 kg)(6.1 × 10^7 m/s)^2/(-1.6 × 10^-19 C)(0.65 T)

Simplifying the expression, we find that the potential difference required is approximately 88.6 volts.

Answer:

19.6 m

Explanation:

The work-energy theorem applies here,

The theorem states that the change in momentum of a particle between two points is equal to the work done in moving the force between the two distance.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Kinetic energy = (1/2)(m)(v²)

m = 3.5 kg, v = 1.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(1.6²) = 4.48 J

ΔK.E = - 4.48 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement.

W = - Fd = - 1.6 F

ΔK.E = W

- 4.48 = - 1.6 F

F = 2.8 N

when the initial velocity is increased by a factor 3.5,

v = 1.6×3.5 = 5.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(5.6²) = 4.48 J

ΔK.E = - 54.88 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement. The frictional force is the same as above.

W = - Fd = - (2.8) d

ΔK.E = W

- 54.88 = - 2.8 d

d = 19.6 m

Compared to 1.6 m, 19.6 m is an increase by a factor 12.25.

Answer:

1) [tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]

Explanation:

1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:

[tex]K_{A} = K_{B} + U_{g,B}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}[/tex]

[tex]\frac{1}{2} \cdot v_{A}^{2} = \frac{1}{2} \cdot v_{B}^{2} + g \cdot h_{B}[/tex]

The minimum speed is obtained herein:

[tex]v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}[/tex]

[tex]v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}[/tex]

[tex]v_{A} \approx 8.272\,\frac{m}{s}[/tex]

The minimum speed he need at launch point is [tex]8.27m/s[/tex]

Energy is neither be created nor be destroyed just change into one form to another form.

              [tex]\frac{1}{2}mv^{2} =\frac{1}{2}mv_{a}^{2} +mgh\\ \\v=\sqrt{v_{a}^{2}+2gh }[/tex]

Where,

Given that, [tex]v_{a}=6.7m/s,h=1.2m,g=9.8m/s^{2}[/tex]

Substitute all values in above relation.

             [tex]v=\sqrt{(6.7)^{2}+2*9.8*1.2 } \\\\v=\sqrt{68.41}=8.27m/s[/tex]

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Answer: 10^-3 V^2/Hz

Explanation:

1 Hz:

Su(f) = No * |H(f)|^2

= 10^-3 * 1/(1+(2*pi*f*R*C)^2)

= 10^-3 V^2/Hz

Answer:

Explanation:

1 )

Charge density of left plate

= - Q / A

Charge density of right plate

= + Q / A

2 )

capacitance c = ε₀ A / d

potential difference = charge / capacitance

= Q / [ ε₀ A / d ]

= Q d  / ε₀ A

Final answer:

The charge density on each plate is ±Q/A, the electric potential difference between the plates is calculated using the electric field and the separation distance, and the capacitance C=ε₀A/d demonstrates that the size of the plates does not affect their capacitance as long as their proportion remains constant.

Explanation:

When considering two parallel plates each with area A and charges of +Q and -Q respectively, separated by a distance d, we can address the posed questions systematically.

Finding the Charge Density on the Plates

The surface charge density σ is defined as charge per unit area. Given the total charge +Q or -Q and the area A of each plate, the charge density on each plate is σ = ±Q/A. This is a direct result of the uniform distribution assumption of the charges across the plates.

Finding the Electric Potential Difference Between the Plates

The electric field E created between the plates by the charge distribution is uniform and can be represented as E = σ/ε₀, where σ is the surface charge density and ε₀ is the vacuum permittivity. Consequently, the electric potential difference V between the plates can be derived from the relation V = Ed, linking the electric field and the separation of the plates.

Demonstrating the Capacitance of Enlarged Plates Remains Constant

The capacitance C of a parallel-plate capacitor is given by C = ε₀A/d, which is independent of the charge on the plates. This formula illustrates that the capacitance is solely dependent on the physical characteristics of the capacitor (i.e., the area of the plates A, the distance between them d, and the permittivity of free space ε₀), and does not change with the amount of charge nor with the size of the plates as long as their proportional relationship remains constant.

Answer:

2s

Explanation:

The position function of the motion can be expressed as:

[tex]s = s_0 + v_0t + at^2/2[/tex]

where [tex]s_0 = <0,0,0>[/tex] is the origin where the particle starts off, [tex]v_0 = <0,0,2> m/s[/tex] and a = <1,2,0> m/s2. In the 3-coordinate system it can be written as

[tex]s = <0 + 0t+ t^2/2, 0 + 0t + 2t^2/2, 0 + 2t + 0t^2/2> = <t^2/2, t^2, 2t>[/tex]

For the particle to hit the 2x+y-z=4 plane then its coordinates must meet that criteria

[tex]2t^2/2 +t^2-2t = 4[/tex]

[tex]2t^2 - 2t -4 = 0[/tex]

[tex]t^2 - t - 2 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{1\pm \sqrt{(-1)^2 - 4*(1)*(-2)}}{2*(1)}[/tex]

[tex]t= \frac{1\pm3}{2}[/tex]

t = 2 or t = -1

Since t can only be positive we will pick t = 2s

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals [tex]F_{Friction}[/tex] = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

Answer:

a)   E = -4.8 10⁻⁵ N / C , b) E = 0 , c) E = 0

Explanation:

For this exercise let's use Gauss's law

         Ф = E. dA = [tex]q_{int}[/tex] / ε₀

As a Gaussian surface we use a sphere, whereby the electric field lines are parallel to the normal area, and the scalar product is reduced to the algebraic product

a) the field for r = 30 cm

At this point we place our Gaussian surface and see what charge there is inside, which is the charge of the solid sphere (r> 20cm), the charge on the outside does not contribute to the flow

           E = q_{int} / A ε₀

           q_{int} = -4.8 10-16 C

The area of ​​a sphere is

              A = 4π R²

We replace

             E = -4.8 10⁻¹⁶ / 4π 0.30² 8.85 10⁻¹²

             E = -4.8 10⁻⁵ N / C

b) r = 45 cm

 This point is inside the spherical conductor shell, as in an electric conductor in electrostatic equilibrium the charges are outside inside the shell there is no charge for which the field is zero

        E = 0

c) R = 60 cm

This part is outside the two surfaces

   The chare inside is

            q_{int} = -4.8 10⁻¹⁶ + 4.8 10⁻¹⁶

            q_{int} = 0

Therefore the electric field is

         E = 0

Answer:

T = 600K

Explanation:

See attachment below.

Answer:

The required angular velocity (ω) will be [tex]0.313~rads^{-1}[/tex].

Explanation:

Due to the rotation of the space station the astronauts experience a centripetal acceleration towards the centre of the space station. If '[tex]\large{a_{c}}[/tex]', 'ω' and 'R' represent the centripetal acceleration, angular velocity of the space station and the radius of the space station respectively, then

[tex]a_{c} = \omega^{2}.R[/tex]

As according to the problem the space station has to rotate in such an angular velocity that it produces the same "artificial gravity" as Earth's surface, we can write

[tex]a_{c} = g = 9.8 ms^{-2}[/tex]

Also given [tex]R = \dfrac{diameter~of~the~space~station}{2} = \dfrac{200 m}{2} = 100 m[/tex]

Therefore we can write,

[tex]&& a_{c} = g = \omega^{2}.R\\&or,& \omega = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{9.8 ms^{-1}}{100 m}} = 0.313~rads^{-1}[/tex]

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 [tex]KV_{LL}[/tex]

(c) The sending-end voltage for 0.9 leading power factor is 33.40 [tex]KV_{LL}[/tex]

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

Answer: e

Explanation:

Argument against the person, circumstantial.

The given question is incomplete. The complete question is as follows.

Two hollow conducting spheres (radius r) with a uniformly distributed charge are placed a distance d apart center to center. A thin wire with a switch S is connected to the surface of each sphere. The switch is initially open.

a. What is the potential between points a and b?

b. If the switch is then closed, what is the charge on each sphere at time [tex]t \rightarrow \infty[/tex].

c. What is the potential between points a and b after the sphere reaches its steady state?

Explanation:

(a) In order to bring a positive test charge from infinity to a point 'a', the work done is equal to the potential energy of the charge at point 'a'.

Hence,      [tex]V_{a} = \frac{1}{4 \pi \epsilon_{o}} \frac{q}{a}[/tex]

Now, work done in bringing a positive test charge from infinity to point 'b' is equal to the potential energy of the charge at point 'b'.

      [tex]V_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{-q}{b}[/tex]      

     [tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]

Therefore, the potential between points a and b is as follows.

  [tex]V_{a} - V_{b} = \frac{q}{4 \pi \epsilon_{o}}(\frac{1}{a} + \frac{1}{b})[/tex]

(b)   As the spheres are connected through a conducting wire then charges will flow from one sphere to another unless and until the charge on both the sphere will become equal. In this case, it is equal to zero.

(c)   Since, the charge of both the spheres is equal to zero so, no work is necessary to bring another charge to a and b. Therefore, potential difference between the points will also become equal to zero.    

Answer:

[tex]6.65\times 10^5 m[/tex]

Explanation:

We are given that

Magnetic field=B=[tex]4\times 10^{-8} T[/tex]

[tex]v=2.7\times 10^7 m/s[/tex]

We have to find the height of proton from the surface of the Earth.

Mass of proton,[tex]m_p=1.67\times 10^{-27} kg[/tex]

Charge on proton,[tex]q=1.6\times 10^{-19} C[/tex]

Radius of Earth, r=[tex]6.38\times 10^6 m[/tex]

Centripetal force due to rotation of proton=[tex]\frac{mv^2}{r+h}[/tex]

Magnetic force,F=[tex]qvB[/tex]

[tex]\frac{mv^2}{r+h}=qvB[/tex]

[tex]\frac{mv}{r+h}=qB[/tex]

Substitute the values

[tex]\frac{1.67\times 10^{-27}\times (2.7\times 10^7)}{6.38\times 10^6+h}=1.6\times 10^{-19}\times 4\times 10^{-8}[/tex]

[tex]6.38\times10^6+h=\frac{1.67\times 10^{-27}\times 2.7\times 10^7}{1.6\times 10^{-19}\times 4\times 10^{-8}}[/tex]

[tex]6.38\times10^6+h=7.045\times 10^6[/tex]

[tex]h=7.045\times 10^6-6.38\times 10^6[/tex]

[tex]h=0.665\times 10^6=6.65\times 10^5 m[/tex]

A charged particle is injected into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, the particle will follow a circular path.

Option D

Explanation:

Magnetic force causes charged particles to move in spiral paths. The Particle accelerators keep the protons to follow circular paths when it is in the magnetic field. Velocity has a change in direction but magnitude remains the same when this condition exists.

The magnetic force exerted on the charged particle is given by the formula:

         [tex]F=q v B \sin \theta[/tex]

where

q is the charge

v is the velocity of the particle

B is the magnetic field

[tex]\theta[/tex] is the angle

In this problem, the velocity is perpendicular to the magnetic field vector, hence

[tex]\theta[/tex] = [tex]90^{\circ}[/tex] and sin[tex]\boldsymbol{\theta}[/tex] =sin 90 degree = 1.

So applying the formula,

the force is simply [tex]F=q v B[/tex]

Also, the force is perpendicular to both B and v and so according to the right-hand rule, we have:

This means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion.

Answer:

Explanation:

Minimum intensity occurs due to destructive interference of sound. For it to take place ,

path difference = odd multiple of half wave length

When moved by .025 m ,

path difference created = ( x + .25 ) - ( x - .25 )

= 2 x .25

= .5 m

So .5 = half wave length

wave length = 2 x .5

= 1 m

frequency = velocity / wave length

= 343 / 1

= 343 Hz

B )

Now when frequency is increased , wave length will be decreased . For maximum intensity , constructive interference will have to take place .

For that

path difference = integral multiple of wave length

0.5 = 1 x wave length

wave length = .5

frequency = 343 / .5

= 686 Hz

Answer: a) vox = vo × cos θ, b) voy =vo× sin θ,

c) H=2.94 m, d) t = vo sinθ / g, e) R = 38.57 m

Explanation:

A)

The velocity v0 is at angle θ to the horizontal.

The horizontal component of vo (vox), vo and the vertical component of vo (voy) all form a right angle triangle.

With vo as the hypotenus, vox as the adjacent and voy as the opposite.

To get vox, we relate vo and vox ( hypotenus and adjacent)

From trigonometry

Cos θ relates hypotenus and adjacent, hence we have that

Cos θ = vox/vo

vox = vo × cos θ

B)

To get the vertical component of vo, we relate vo and voy ( hypotenus and opposite).

According to trigonometry, sin θ relates hypotenus and opposite, hence we have that

Sin θ = voy/vo

voy =vo× sin θ

C)

The formulae for the maximum height of a projectile motion is given as

H = vo² (sin θ)²/2g

Where g = acceleration due to gravity = 9.8 m/s²

By substituting the parameters, we have that

H = 26² × (sin 17)²/2(9.8)

H = 676 × 0.0854/19.6

H = 57.7304/ 19.6

H = 2.94 m

D)

This is the motion of a projectile and the conditions at maximum height are vy = 0 and ay = - g

From the equation of motion

vy = voy - gt

0 = voy - gt

But voy = vo sinθ

0 = vo sinθ - gt

gt = vo sinθ

t = vo sinθ / g

E)

The horizontal distance covered formulae is given by

R = u² sin2θ/g

R = 26² × sin 2(17)/9.8

R = 676 × sin 34/ 9.8

R = 378.014/ 9.8

R = 38.57 m

The correct Answer is:

Learn more about:

https://brainly.com/question/19028701

The electromagnetic force generates the heat and light produced by the sun and other stars.

The force that generates the heat and light produced by the sun and other stars is the electromagnetic force. This force is responsible for holding atoms together and producing electromagnetic radiation, which includes heat and light. It is much stronger compared to the weak force and gravity.

Answer:

240 m

120 m

Explanation:

d = Path difference = 120 m

For destructive interference

Path difference

[tex]d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m[/tex]

The longest wavelength is 240 m

For constructive interference

[tex]d=\lambda\\\Rightarrow 120\ m=\lambda[/tex]

The longest wavelength is 120 m

Final answer:

The longest wavelength for destructive interference at point Q, where the path difference is 120 m, is 240 m.

Explanation:

The longest wavelength for which there will be destructive interference at point Q can be found by considering the path difference between the waves from the two antennas at point Q. The waves must differ by an odd multiple of ½ wavelengths for destructive interference to occur. The distance from antenna A to point Q is 120 m + 40 m = 160 m, and the distance from antenna B to point Q is 40 m. Therefore, the path difference is 120 m.

For the first occurrence of destructive interference, the path difference should be ½ wavelength, so we calculate the longest wavelength (λ) as:

½ λ = 120 m

λ = 240 m

This is the longest wavelength that causes destructive interference at point Q.

Answer:

a) [tex]v(2) = 3.9\,\frac{m}{s}[/tex], b) [tex]v(4) = -15.7\,\frac{m}{s}[/tex]

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

[tex]v(t) = 23.5 -9.8\cdot t[/tex]

The velocities at given instants are, respectivelly:

[tex]v(2) = 3.9\,\frac{m}{s}[/tex]

[tex]v(4) = -15.7\,\frac{m}{s}[/tex]