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2 years ago

7

What is the total resistance in this circuit?

1 answer:

VikaD [51]2 years ago

6 0

Try 40. it seems correct. i’m sorry if i’m wrong.

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What are particles of carbon called?

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3 years ago

Mary and her younger brother Alex decide to ride the 26 ft diameter carousel at the State Fair. Mary sits on one of the horses i

hammer [34]

Answer:

$a_M=1.92a_A$

Explanation:

$\omega_M=\omega_A$ = Angular speed

$r_M$ = Distance of Mary = 11.5 ft

$r_A$ = Distance of Alex = 6 ft

Ratio of centripetal acceleration is given by

$\dfrac{a_M}{a_A}=\dfrac{\omega_M^2r_M}{\omega_A^2r_A}\\\Rightarrow \dfrac{a_M}{a_A}=\dfrac{r_M}{r_A}\\\Rightarrow a_M=a_A\dfrac{r_M}{r_A}\\\Rightarrow a_M=\dfrac{11.5}{6}a_A\\\Rightarrow a_M=1.92a_A$

Mary's centripetal acceleration is 1.92 times the centripetal acceleration of Alex

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3 years ago

In which of these cases is a waiter doing work on the object

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2 years ago

Which statement correctly stated kepler's 3rd law of planetary motion?

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3 years ago

Read 2 more answers

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th

lozanna [386]

Answer:

$k=105.0359\times 10^4\,W.m^{-1}.K^{-1}$

Explanation:

Given:

temperature at the hotter end, $T_H=100^{\circ}C$

temperature at the cooler end, $T_C=0^{\circ}C$

length of rod through which the heat travels, $dx=0.7\,m$

cross-sectional area of rod, $A=1.1\times 10^{-4}\,cm^2$

mass of ice melted at zero degree Celsius, $m=8.7\times 10^{-3}\,kg$

time taken for the melting of ice, $t=15\times60=900\,s$

thermal conductivity k=?

By Fourier's Law of conduction we have:

$\dot{Q}=k.A.\frac{dT}{dx}$ ......................................(1)

where:

$\dot{Q}$ =rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice, $L = 334000\,J.kg^{-1}$

$Q=m.L$

$Q=8.7\times 10^{-3}\times 334000$

$Q=2905.8\,J$

Now the heat rate:

$\dot{Q}=\frac{Q}{t}$

$\dot{Q}=\frac{2905.8}{900}$

$\dot{Q}=3.2287\,W$

Now using eq,(1)

$3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}$

$k=205.4606\,W.m^{-1}.K^{-1}$

8 0

3 years ago