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jek_recluse [69]
3 years ago
10

A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.

If the rails are separated by 20 cm, what is the power generated by the rod? O 341 x 1ow O 6.98 x 103 w O 9.00x 10sw O 8.10 x 105w
Physics
1 answer:
love history [14]3 years ago
3 0

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

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geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0
1 year ago
7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2
alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

\displaystyle a = \frac{\Sigma F}{m},

where

  • a is the acceleration of the object in \text{m}\cdot\text{s}^{-2},
  • \Sigma F is the net force on the object in Newtons, and
  • m is the mass of the object in kilograms.

As a result,

\displaystyle m = \frac{\Sigma F}{a}.

Assume that all other forces on this object are balanced. The net force on the object will be 100\;\text{N}. The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}.

\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}.

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}.

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Answer

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Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

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\eta =26.03 %

b) Work done by the engine

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   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

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