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Ket [755]
3 years ago
5

4. A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exer

ted downward at an angle of 35.2° below the horizontal. Find mk between the box and the floor.
Physics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

0.61°

Explanation:

Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.

Pulling force= resistance force

From the formula for pulling force,

F(x)= Fcos(θ)

= 425×cos(35.2)

=347N

The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)

=425×0.567=245N

Resistance force= (325N+ 245N) (α)= 570N(α)

We can now equates the pulling force to resistance force

570 (α)= 347N

(α)= 347/570

= 0.61

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The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that

4 m/s = <em>a</em> (2.5 s)   →   <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude <em>F</em> such that

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A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy
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Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

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Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

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The magnitude of the magnetic field is given by:

B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B =  \frac{4\pi * 10^{-7}   * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }

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3 years ago
A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows
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Answer:

(a) The impedance in the circuit is Z=183.33\Omega.

(b)The resistance is R=38.89\Omega.

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

Z=\frac{V_rms}{I_rms}

Here, V_rms is the rms voltage and I_rms is the rms current.

PutV_rms=110 V and I_rms=0.600 A.

Z=\frac{110}{0.600}

Z=183.33\Omega

Therefore, the impedance in the circuit is Z=183.33\Omega.

(b)

The expression for the average power is as follows;

P_{a}=I_{rms}^{2}R

Here, P_{a} is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

R=\frac{P_{a}}{I_{rms}^{2}}

Put P_{a}=14W and

R=\frac{14}{{0.600}^{2}}

R=38.89\Omega

Therefore, the resistance is R=38.89\Omega.

(c)

The expression for the impedance is as follows;

Z^{2}=R^{2}+X_{L}^{2}

Here,X_{L} is the inductive reactance.

Put Z=183.33\Omega and R=38.89\Omega.

(183.33)^{2}=(38.89)^{2}+X_{L}^{2}

X_{L}=179.16\Omega

The expression for the inductive reactance in terms of  frequency is as follows;

X_{L}=2\pi fL

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

L=\frac{X_{L}}{2\pi f}

Put X_{L}=179.16\Omega and f=50Hz.

L=\frac{179.16}{2\pi (50)}

L=0.57 H

Therefore, the inuctance is 0.57 H.

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