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Brut [27]
3 years ago
15

A small metal block with mass of 140 g sits on a horizontally rotating turntable. The turntable makes exactly 2 revolutions each

second. The disk is located 6 cm from the axis of ration of the turntable. Answer each of the following question:
a. What is the frictional force acting on the disk?
b. The disk will slide off the turntable If it is located at a radius larger than 11 cm from the axis of rotation. What is the coefficient of static friction?
Physics
1 answer:
andre [41]3 years ago
3 0

Answer:

a) Frictional force acting on the disc = 1.326N

b) Coefficient of static friction = 1.77

Explanation:

mass of the metal block = 140 g = 0.14 kg

Frequency of the turn table = 2 Revs/sec

distance of the disc from the axis of the turn table, r = 6 cm = 0.06 m

a) Frictional force acting on the disc

F = ma

a = V²/r ; V = ωr;

a = ω²r²/r    ;   a = ω²r

F = mω²r

ω = 2πf = 2π * 2 = 4π

ω =  4π

F = 0.14 * (4π)² * 0.06

F = 1.326 N

b) Coefficient of static friction

F = μmg.............(1)

F = mω²r...........(2)

Equating (1) and (2)

μmg =  mω²r

μ = ω²r/g

r = 11 cm = 0.11 m

g = 9.8 m/s²

μ = ( 4π)²  *  0.11/9.8

μ = 1.77

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solong [7]

Answer:

Before the plank will tip cat will walk 1.652 m

Explanation:

Mass of the cat along with plank m_1=7kg

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We have to find how far right of sawhorse B.

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Balancing the the torque

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7\times 9.8\times 0.850=3.6\times 9.8\times d_2

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5 0
3 years ago
A coil formed by wrapping 65 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the
Fiesta28 [93]

Answer:

377 m

Explanation:

number of turns, N = 65

θ = 36°

B1 = 200 micro Tesla

B2 = 600 micro tesla

t = 0.4 s

induced emf, e = 80 mV

Let a be the side of the square coil.

e=\frac{d\phi }{dt}=NA\frac{dB}{dt}\times Sinθ

0.080=\frac{65\times a^{2}\times Sin36\times\left ( 600 - 200 \right )\times 10^{-6}}{0.4}

0.080=0.038a^{2}

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Total length of the wire, L = N x 4a = 65 x 4 x 1.45 = 377 m

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7 0
4 years ago
In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has
irinina [24]

Answer:

 Coil 2 have  235  loops

Explanation:

Given  

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is  ε ₁ = 2.78 V

The emf induced in coil 2 is  ε ₂ = 4.11 V

Let

n ₂  is the number of loops in coil 2.

Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂=\frac{159 * 4.11}{2.78}

n₂=235

Therefore, the coil 2 has  n ₂= 235  loops.

4 0
3 years ago
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Imagine that a hypothetical life form is discovered on our moon and transported to Earth. On a hot day, this life form begins to
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4 0
3 years ago
8
Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along  positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

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Now, since v_1\ and\ v_2 are perpendicular to each other, their resultant magnitude is given as:

|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}

Plug in the given values and solve for the magnitude of the resultant.This gives,

|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0
3 years ago
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