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sp2606 [1]
3 years ago
10

suppose that the 25 resistor in problem 8 is replaced by a 45 resistor. without performing any calculations, describe qualitativ

ely the change in each of the following

Physics
1 answer:
Natasha2012 [34]3 years ago
5 0
A) Equivalent resistance increases
b) Current through entire circuit decreases
c) Current through the two ( 55 and 75 resistors remains the same), current through 45-resistor is 0.2A
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Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
3 years ago
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