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4 years ago

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For the magnetic field at some random angle to the plane of the small coil, draw a picture showing only the small coil, a vector

giving the direction of the magnetic field, the area vector for the coil, and the angle between the magnetic field and the area vector. Write an equation for the magnetic flux through the small coil at an instant of time when the area vector is at some angle to the magnetic field. Write an expression that shows how the magnetic flux through the small coil changes as it turns with a constant angular speed.

1 answer:

bearhunter [10]4 years ago

3 0

Answer:

Solution attached below

Explanation:

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A student is holding a stone at a certain height. The stone has 50 joules of potential energy and 0 joules of kinetic energy. Th

Hitman42 [59]

Answer:

d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy .

Explanation:

Given :

Initial Potential energy , $P_i=50\ J$ .

Initial Kinetic energy , $K_i=0\ J$ . ( because ball is in rest )

Now , we know , kinetic energy is maximum when an object reaches ground .

Also , potential energy is zero when an object is in ground .

We know , by conservation of energy :

Initial total energy = Final total energy

$P_i+K_i=P_f+K_f\\\\50+0=0+K_f\\\\K_f=50 \ J$

Therefore , option d) is correct .

6 0

4 years ago

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

Serhud [2]

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

5 0

3 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

3 0

2 years ago

(PLEASE HELP) What area of the Earth's mantle is the densest?

Nady [450]

The area nearest to the earth's core is the densest..
So your answer would be letter choice ( D ) . . .

Hope it Helped :)

7 0

4 years ago

Read 2 more answers

a kid is on a skate board going 14kph and trows a set of keys on the ground at 8kph. the spped of the keys relitive to the gound

Vanyuwa [196]

To determine the speed relative to the ground, since the ground is our reference frame, it would be v = 0, for the kid on the skateboard, you would need to take into account the speed that he/she is going and the speed of the keys thrown at.

I believe.

5 0

3 years ago