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Basile [38]
3 years ago
9

the structure shown in the picture is made of copper. The three thin rods are of equal dimensions. The single main rod is heated

to 400 °C and then is brought into contact with the rods P, Q, and R. After several minutes, the temperatures of rods P, Q, and R are recorded. Which option gives the correct values of the final temperatures of rods P, Q, and R? A) P = R < Q B) P = Q = R C) P = R > Q D) P = Q < R
Physics
2 answers:
charle [14.2K]3 years ago
4 0

If you are on USA Test prep the answer is B

julia-pushkina [17]3 years ago
3 0

The answer is B) P = Q = R

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Serga [27]
The amount of matter in an object is called its, mass. How much the mass weights would be referred to as weight
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2 years ago
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The truck in which
sertanlavr [38]

Answer:

m = 4

Explanation:

We have,

You apply a force of  600 N to the branch  which acts as a lever. It means it is input force, IF = 600 N

The rear of the truck  weighs 2,400 N. It means it is output force, OF = 2400 N

The ratio of output force to the input force is equal to the mechanical advantage of the lever arm. It is given by :

m=\dfrac{2400}{600}\\\\m=4

So, the mechanical  gain of the lever arm is 4.

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2 years ago
Water is discharged from a pipeline at a velocity v (in ft/sec) given by v = 1306p(1/2), where p is the pressure (in psi). If th
shepuryov [24]

Answer:

a=38.5 ft/sec^{2}

Explanation:

Note that acceleration is the rate change of velocity i.e

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\frac{dv}{dt}=1306*(1/2)p^{-1/2}\frac{dp}{dt} \\

if we substitute value for the pressure as giving in the question and also since the rate change of pressure is 0.354psi/sec, we have

a=653*0.1667*0.354\\

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3 0
2 years ago
Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

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NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

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                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

The concentration of Br^- = 0.1 + s

Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

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Answer:

what  is it on? like name one of the questions

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