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victus00 [196]
2 years ago
7

A metal block of mass 235 g rests at a point 2.8 m from the center of a horizontal rotating wooden platform. The coefficient of

static friction between the block and the platform is 0.437. The platform initially rotates very slowly but the rotation rate is gradually increasing. The acceleration of gravity is 9.8 m/s 2 . At what minimum angular velocity of the platform would the block slide away? Answer in units of rad/s.
Physics
1 answer:
Nuetrik [128]2 years ago
5 0

Answer:

ω=1.23 rad/sec

Explanation:

The block will start sliding when the centrifugal force exceeds the force of friction

We have Force of friction and centrifugal force are as under

F_{f}=\mu mg...........(i)\\\\F_{c.f}=m\omega ^{2}r................(ii)

Thus equating the 2 forces to get the minimum angular velocity we have

\fr\mu mg= m\omega^{2}r\\\\\therefore \omega =(\frac{\mu g}{r})^{1/2}

Applying values we get

ω=1.23 rad/sec

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In this circuit (see picture), which resistor will draw the least power?
Basile [38]
A few different ways to do this: 

Way #1: 
The current in the series loop is  (12 V) / (total resistance) . 
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is  (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power.  That's  R1 .

And by the way, it's not "drawing" the most power.  It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

                 (voltage across the resistance)²  /  (resistance)  .

In a series loop, the voltage across each resistor is

          [ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

         (individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .                                      

Way #3:  (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===>  When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
3 0
3 years ago
The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of
kolbaska11 [484]

Answer:

a) A = 4.0 m , b)   w = 3.0 rad / s , c)  f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

          x = A cos (wt + fi)

In the exercise we are told that the expression is

          x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

         A = 4.0 m

b) the frequency or angular velocity

         w = 3.0 rad / s

c) angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 3 / 2π

           f = 0.477 Hz

d) the period

frequency and period are related

           T = 1 / f

           T = 1 / 0.477

           T = 20.94 s

e) the phase constant

          Ф = 0.10 rad

f) velocity is defined by

          v = dx / dt

         

         v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

         v = A w

          v = 4 3

          v = 12 m / s

g) the angular velocity is

          w² = k / m

          k = m w²

          k = 1.2 3²

          k = 10.8 N / m

h) the total energy of the oscillator is

          Em = ½ k A²

           Em = ½ 10.8 4²

          Em = 43.2 J

i) the potential energy is

           Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

               x = 3.98 m

j) kinetic energy

           K = ½ m v²

for t = 00.1 ²

    v = A w sin 0.10

    v = 4 3 sin 0.10

    v = 1.98 m / s

3 0
3 years ago
What os the konectic energy of 620.0kg coaster moving with a velocity of 9.00m/s
luda_lava [24]
Hope this helps you.

4 0
3 years ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
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The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
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In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
3 years ago
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Dvinal [7]

Answer:

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Explanation:

8 0
3 years ago
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