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4 years ago

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A solid sphere is released from height h from the top of an incline making an angle θ with the horizontal. (a) Calculate the spe

ed of the sphere when it reaches the bottom of the incline in the case that it rolls without slipping. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

1 answer:

Natali [406]4 years ago

5 0

Answer:

v =√ (gh/.700)

Explanation:

As per the conservation of mechanical energy,

( K+ U)i = ( K+U)f

where i is the initial energy and f is the final energy.

Kinetic energy, K(i) at top = 0

when call will reach the bottom , U(f) = 0

U(i) = mgh

K(f) = 1/2 (mv²) + 1/2 I ω²

For solid sphere , I = 2/5 m R²

and ω = v/R

so , K = 1/2 mv² + 1/2*2/5*m*R^2*v^2/R^2

K = .700 mv²

now .700mv² = mgh

v =√ (gh/.700)

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Determine the inductance L of a 0.65-m-long air-filled solenoid 3.2 cm in diameter containing 8400 loops.

den301095 [7]

Answer:

The inductance is $L = 0.1097 \ H$

Explanation:

From the question we are told that

The length is $l = 0.65 \ m$

The diameter is $d = 3.2 cm = 0.032 \ m$

The number of loops is $N = 8400$

Generally the radius is evaluated as

$r = \frac{ 0.032 }{2} = 0.016 \ m$

The inductance is mathematically represented as

$L = \frac{ \mu_o * N^2 * A }{ l }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^{-7} N/A^2$

A is the cross-sectional area which is mathematically evaluated as

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=> $A = 3.142 * (0.016)^2$

=> $A = 0.000804 \ m^2$

=> $L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }$

=> $L = 0.1097 \ H$

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3 years ago

when the armature of an ac generatr rotates at 15.0 rad/s, the amplitude of the induced emf is 27.0 V. What is the amplitude of

nata0808 [166]

To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

$E = \omega NBA |sin \omega t|$

Here,

$\omega$ = Angular frequency

N = Number of turns

B = Magnetic field

The emf has its maximum value when $sin \omega t = \pm 1$

Thus the amplitude of the emf is

$E = \omega NBA$

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

$E \propto \omega$

Further it can be written as follows,

$\frac{E_1}{E_2} \propto \frac{\omega_1}{\omega_2}$

$E_2 = \frac{\omega_2}{\omega_1}E_1$

$E_2 = \frac{10rad/s}{15rad/s}(27.0V)$

$E_2 = 18V$

Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V

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3 years ago

10 points please somebody help me

Anuta_ua [19.1K]

It is the first one?

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3 years ago

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Suppose a candle is burning, and wind is blowing on one side of the flame. Which principle explains why the flame bends toward t

Mekhanik [1.2K]

When a candle is burning and wind is blowing it on one side of the flame, which causes the flame to bend towards the wind is an example of Bernoulli’s principle. The principle explain that the higher the speed, the lower the pressure becomes. When you blow against one side of the flame, you are creating an area of low pressure. The relatively high-pressure air on the other side of the candle will rush over to fill the area of low pressure that causes the flame to be pushed in the direction of the blowing.

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Amy is in-line skating. Her mass is 50 kg. She is rolling forward (north) on a flat section of road at 10 m/s. The rolling frict

Licemer1 [7]

Answer:

The right solution is "0.50 m/s²". A further explanation is provided below.

Explanation:

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m = 50 kg

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On substituting the given values, we get

⇒ $=-10-15$

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⇒ $a = \frac{F}{m}$

On substituting the given values, we get

⇒ $=\frac{-25}{50}$

⇒ $=-0.50 \ m/s^2$

The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.

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3 years ago