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What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.

Answers

Answer 1
Answer:

Answer:

 λ = -47 nC / m  

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.  "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

                                      V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

 λ : The line charge density

- Re-arrange and solve for λ:

                                     λ = V / 2*k*Ln( D / L )

Plug in the values:

                                    λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

                                    λ = -4.74022*10^-8 C / m

                                   λ = -47 nC / m  
Answer 2
Answer:

Final answer:

Line charge density is the total charge distributed along the length of a wire, expressed in coulombs per meter. To calculate it, divide the total charge by the total length of the wire. Without specific numbers for charge and length, a numerical value can't be given.

Explanation:

To calculate the magnitude of the line charge density of a power line, you need to know the total charge (Q) distributed along the total length (L) of the wire. The line charge density (λ) is then defined as λ = Q/L. Unfortunately, without any specific numbers provided for these parameters, I can't provide a numerical answer.

Line charge density is a significant concept in electromagnetism and is measured in coulombs per meter (C/m).

Remember that the charge can be uniform or non-uniform along the length of the line.

For example, if a power line has a total charge of 0.02 C spread along its length of 50 m, it would have a line charge density of λ = Q/L = 0.02 C / 50 m = 0.0004 C/m

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A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer:

5.78971 m

Explanation:

P_1 = Initial pressure = 0.873 atm

P_2 = Final pressure = 0.0282 atm

V_1 = Initial volume

V_2 = Final volume

r_1 = Initial radius = 16.2 m

r_2 = Final radius

Volume is given by

(4)/(3)\pi r^3

From the ideal gas law we have the relation

(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m

The radius of balloon at lift off is 5.78971 m

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of the atom, but they are not in the correct order.

Answers

Answer:

Y, X, Z, W

Explanation:

Jerome must put the given models in the order Y, X, Z, W to display the development of atom from the earliest to the most recent one. 'Y' represents 'Thomson's plum pudding model' came in 1904 which was followed by the 'Rutherford's nuclear atomic model' of 1911 as represented by X. This was succeeded by the 'Bohr's electrostatic model' in 1913(as shown in model Z) and lastly, the model W which exemplifies the 'Quantum Mechanical Model' by Edwin Schordinger in 1926. Thus, the correct order is Y, X, Z, W.

Answer:YXZW

Explanation:

A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answers

Answer:

The moment of inertia of the system is  I = 400.5 \ kg \cdot m^2

Explanation:

From the question we are told that

    The mass of the platform is  m = 133\ kg

     The  radius of the  platform is  r = 1.95 m

     The mass of the person is m_p = 62.7 \ kg

     The position of the person from the center is  d = 1.19 \ m

       The mass of the dog is m_D = 28.5 \ kg

     The position of the dog from the center is  D = 1.45 \ m

   

Generally the moment of inertia of the platform with respect to its axis is  mathematically represented as

       I_p = (m r^2)/(2)

The  moment of inertia of the person with respect to the axis is mathematically represented as

        I_z = m_p* d^2

The  moment of inertia of the dog with respect to the axis is mathematically represented as

       I_D = m_d * D^2

So the moment of inertia of the system about the axis  is mathematically evaluated as

        I = I_p + I_z + I_D

=>      I = (mr^2)/(2) + m_p * d^2 + m_d * D^2

substituting values  

            I = ((133) * (1.95)^2)/(2) + (62.7) * (1.19)^2 + (28.5) * (1.45)^2

          I = 400.5 \ kg \cdot m^2

A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\nor\nP=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_(1)*v_(1))+F*t=(m_(1)*v_(2))

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

(10*1)+F*5=10*4\n10+5*F=40\n5*F=40-10\n5*F=30\nF=6[N]

A long wire carries a current toward the south in a magnetic field that is directed vertically upward. What is the direction of the magnetic force on the wire?

Answers

Answer:

the answer is it is going north

Explanation:

because its the opposite

Final answer:

The magnetic force on a wire carrying current towards the south under a magnetic field directed vertically upwards will point towards the East. In order to determine this, use the right-hand rule.

Explanation:

The direction of the magnetic force on a current-carrying wire under a magnetic field can be deduced using the right-hand rule. In this case, with the current flowing towards the south and the magnetic field directed vertically upward, you would point your right thumb in the direction of the current (southwards) and curl your fingers in the direction of the magnetic field (upwards). The palm of your hand will then face toward the direction of the force. In this case, the force would be pointing toward the East.

The right-hand rule is a vital principle in the study of electromagnetism as it aids in identifying the direction of various quantities in magnetic fields.  The magnetic force on a current-carrying wire represents the phenomenon underlying the working of many electric motors.

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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m'

Answers

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

(c) 8.72 m/s, upward

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

t=(1.77 s)/(2)=0.89 s

So we can now solve the equation for u, and we find:

u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

v^2 - u^2 = 2ad

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

d=(v^2-u^2)/(2a)=(0^2-(8.72 m/s)^2)/(2(-9.8 m/s^2))=3.88 m
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