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insens350 [35]
3 years ago
7

A small rubber wheel drives the rotation of a larger pottery wheel by running along its edge. The small wheel radius is 1.2 cm,

and it accelerates at 3 rad/s2. The pottery wheel has a radius of 36 cm. What is the angular acceleration of the pottery wheel? How long till the pottery wheel rotates at 60 rpm?
Physics
1 answer:
Lisa [10]3 years ago
5 0

Answer:

α₂= 0.1  rad/s²

t= 62.8 s

Explanation:

Given that

For small wheel

r₁= 1.2 cm

α₁ = 3 rad/s²

For large wheel

r₂= 36 cm

Angular acceleration = α₂  rad/s²

The tangential acceleration for the both wheel will be same

a = α₁ r₁=α₂ r₂

Now by putting the values in the above equation

α₁ r₁=α₂ r₂

3 x 1.2 = 36 x α₂

α₂= 0.1  rad/s²

Given that

N = 60 rpm

Angular speed in rad/s ω

\omega = \dfrac{2\pi N}{60}

\omega = \dfrac{2\pi \times 60}{60}

ω = 6.28 rad/s

Time taken is t

ω = α₂ t

6.28 = 0.1 t

t= 62.8 s

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A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
3 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane
LiRa [457]

Answer:

Explanation:

mass of baseball, m = 0.148 kg

initial velocity, u = 15.5 m/s

final velocity, v = 10.1 m/s

Impulse is defined as the change in momentum of the body.

Impulse = change in momentum

I = m (v - u)

I = 0.148 (10.1 - 15.5)

I = - 0.8 Ns

3 0
3 years ago
The primary colors of light are A. red, yellow, and blue. B. red, white and blue. C. blue, green, and red. D. cyan, magenta, and
zhuklara [117]
I think it is A I hope I helped you
3 0
2 years ago
Read 2 more answers
An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surfac
Harlamova29_29 [7]

Answer:

b. 48.0 g.

Explanation:

Given;

mass of the exoplanet, M_p = 3M_e

radius of the exoplanet, r_p = \frac{1}{4} r_e

The acceleration due to gravity of the planet is calculated as;

g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48  \ g

Therefore, the correct option is b. 48.0 g

5 0
2 years ago
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