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1 month ago

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What is E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the distance r f

rom the axis of the cylindrical rod?

1 answer:

Sav [3.1K]1 month ago

3 0

Complete question:

An infinitely long conducting cylindrical rod carries a positive charge λ per unit length and is enveloped by a conducting cylindrical shell (which is also infinitely long) that has a charge per unit length of −2λ and a radius of r1, as depicted in the illustration.

Answer:

E(r) = λ/2πϵ0r

For further detail, see the attached file.

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The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

ValentinkaMS [3465]

Answer:

The force magnitude is $F_{net}= 1.837 *10^4N$

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, $\theta = 20^o$

The angular rate of increase is $w = 2 \ ^o/s$

Converting to revolutions per second gives us $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The rope length is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , Tension T of the rope is 18 kN.

The weight of the para-sailor is $M_p = 75kg$

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

The derivative of displacement results in velocity.

Hence,

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

signifies the velocity, and further differentiation yields acceleration.

Therefore,

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

$w = \frac{\Delta \theta}{\Delta t}$

where $\theta$ represents the angular displacement.

Next, evaluating the interval from $20^o \ to \ 30^o$ gives us

$2 = \frac{30 -20 }{t -0}$

making t the focal point.

$t = \frac{10}{2}$

$= 5s$

At this time, the displacement measures

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity computes to

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

Whereas linear acceleration calculates as

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally, radial acceleration is given by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Simultaneously, angular acceleration can be represented as

$\alpha_t = r \theta'' + 2 r' \theta '$

Then $\theta '' = \frac{d (0.0349)}{dt} = 0$

Thus,

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The resultant acceleration is mathematically denoted as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the acceleration's direction is mathematically expressed as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The y-axis force acting on the para-sailor is mathematically shown as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The x-axis force acting on the para-sailor is represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The overall force is calculated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The directional force is evaluated as

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

2 months ago

A team of engineering students is testing their newly designed raft in the pool where the diving team practices.

Yuliya22 [3333]

Answer:

Lower than

Explanation:

When the cube is placed on the raft, the displaced water equals the combined weight of both the cube and raft. In contrast, when submerged in water, the displaced water corresponds to the raft's weight plus the cube's volume. Since the cube's volume is smaller than what is needed to displace its weight with water, the water level is lower.

6 0

2 months ago

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

Yuliya22 [3333]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>

5 0

2 months ago

An elevator is being pulled up from the ground floor to the third floor by a cable. The cable is exerting 4500 newtons of force

serg [3582]

The gravitational force acting on the elevator is 4500N. Explanation: The parameters provided indicate that the force exerted by the elevator is F = 4500 N. The elevator does not accelerate. According to Newton's third law, the force exerted by the cable matches the gravitational force on the elevator, which represents its weight (W), and the elevator's motion stands as follows: F = W + (Mass of elevator × Acceleration of elevator). Therefore, F = W + (Mass of elevator × 0) = W, which leads us to F = 4500 N = W. The net force on the elevator can be expressed as F - W = 0, confirming that the gravitational force on the elevator equals W = 4500N.

3 0

1 month ago

If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,

Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0

1 month ago