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3 years ago

12

How much heat is absorbed by a 28g iron skillet when its temperature rises from 7oC to 29oC?

1 answer:

prohojiy [21]3 years ago

5 0

From the information given above,
Mass [M] = 28 g
Change in temperature = 29 - 7 = 22
Specific heat of iron = 0.449 [This value is constant]
The formula for calculating heat absorbed, Q is
Q = Mass * Specific heat of Iron * change in temperature
Q = 28 * 0.449 * 22 = 276.58 J<span />

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If f(x)=4/x+2 and g is the inverse of f,then g'(10)=

ch4aika [34]

Answer:

g'(10) = $\frac{-1}{16}$

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x <em> </em><em>(Identity)</em>

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = $\frac{1}{f'(x)}$ --equation[1] Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = $\frac{4}{x}$ + 2

$\frac{4}{x}$ = 8

x = $\frac{1}{2}$

Since f(x) = $\frac{4}{x}$ + 2

f'(x) = - $\frac{4}{x^{2} }$

f'( $\frac{1}{2}$ ) = -4 × 4 = -16

Putting it in equation 1, we get:

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2 years ago

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

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$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

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