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2 years ago

In the system illustrated by the diagram, the magnetic field is increasing. In which

Physics

1 answer:

Black_prince [1.1K]2 years ago

3 0

The emf will be induced in anti-clockwise direction.

<u>Explanation</u>

Lenz's law tells us the direction us the direction that the current will flow. It states that the direction is always such that it will oppose the change in flux which produced it. This means that any magnetic field produced by an induced current will be in opposite direction to the change in the original field.

To find the direction of emf, Stretch the forefinger, middle finger and the thumb of the right hand mutually perpendicular to each other. If the force finger points in the direction of the magnetic field, the thumb gives the direction of the motion of the conductor then the middle finger gives the direction of the induced current.

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2 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

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Solve for theta:

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1 year ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

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2 years ago

A truck is traveling at 80 m/s and has 12,000 J of kinetic energy. What is the truck’s mass?

Murrr4er [49]

Answer:

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2 years ago

What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are

nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is $15.58 g/cm^3$ .

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = $19.32g/cm^3$

The density of silver = $10.1g/cm^3$

The density of copper = $8.8g/cm^3$

$Volume=\frac{Mass}{Density}$

Volume of the gold in the mixture = $V_1=\frac{18x}{19.32 g/cm^3}$

Volume of the silver in the mixture = $V_2=\frac{5x}{10.1 g/cm^3}$

Volume of the copper in the mixture = $V_3=\frac{1x}{8.8 g/cm^3}$

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = $V_1+V_2+V_3$

Density of the mixture:

$\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3$

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