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kow [346]
3 years ago
15

A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is

Physics
2 answers:
SOVA2 [1]3 years ago
4 0
I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>
icang [17]3 years ago
3 0
Hello there.

<span>A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is

</span><span>4.0 N</span>
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Answer:

it snaps

Explanation:

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3 years ago
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17 C and
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Answer:

Q = - 4312 W = - 4.312 KW

Explanation:

The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:

Q = - kA dt/dx

Integrating from one side of the slab to other along the thickness dimension, we get:

Q = - kA(T₂ - T₁)/L

Q = kA(T₁ - T₂)/t

where,

Q = Rate of Heat Loss = ?

k = thermal conductivity = 1.4 W/m.k

A = Surface Area = (11 m)(8 m) = 88 m²

T₁ = Temperature of Bottom Surface = 10°C

T₂ = Temperature of Top Surface = 17° C

t = Thickness of Slab = 0.2 m

Therefore,

Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m

<u>Q = - 4312 W = - 4.312 KW</u>

<u>Here, negative sign shows the loss of heat.</u>

3 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

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V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

4 0
3 years ago
With pollution on the rise, and demand for fresh water increasing, world leaders are concerned that in the future many people wi
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