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Ivahew [28]
3 years ago
7

A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the r

atio of their angular velocities,ω1 / ω2 ?
Physics
1 answer:
Gwar [14]3 years ago
5 0

Answer:

The  ratio is  \frac{w_1}{w_2}  =  \frac{R_2}{R_1}

Explanation:

From the question we are told that

   The first radius is R_1

    The  second radius is  R_2

   

Generally the angular speed of the turntable is mathematically represented as

       w_1  =  \frac{ v_k }{R_1 }

Generally the angular speed of the rubber roller is mathematically represented as

        w_2 =  \frac{ v_k }{R_2 }

Where v_k is the velocity of both turntable and  rubber roller

So

    \frac{w_1}{w_2}  =  \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }

    \frac{w_1}{w_2}  =  \frac{R_2}{R_1}

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Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

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3 years ago
It is hot outside, and the air conditioning has been turned on inside two buildings. Building A's walls contain insulation, but
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Answer:

building A because the insulation will keep the air in the building while more so building b will let the air flow out because there is no insulation

Explanation:

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3 years ago
What does g mean in physics
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Which one of the following is a correct statement......
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3 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

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4 years ago
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