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Sergio039 [100]
3 years ago
11

A piece of cork (density 250 kg/m3 ) of mass 0.01 kg is held in place under water (density 1000 kg/m3 ) by a string. What is the

tension, T, in the string? [Use g = 10 m/s2 ]
Physics
2 answers:
Vesnalui [34]3 years ago
5 0

Answer:

Tension = 0.3 N

Explanation:

As we know that the cork is inside water

so the buoyancy force on the cork is counter balanced by tension force in string and weight of the block

So the force equation is given as

F_b = T + mg

now we will have

Volume = \frac{mass}{density}

V = \frac{0.01}{250} = 4 \times 10^{-5} m^3

now buoyancy force on the block is given by

F_b = \rho V g

F_b = 1000(4 \times 10^{-5})(10)

F_b = 0.4 N

now by force balance equation

0.4 = T + 0.01(10)

T = 0.4 - 0.1 = 0.3 N

vekshin13 years ago
3 0

Answer:

0.3 N

Explanation:

mass of cork = 0.01 kg, density of cork = 250 kg/m^3

density of water = 1000 kg/m^3, g = 10 m/s^2

Tension in the rope = Buoyant force acting on the cork - Weight of the cork

Buoyant force = volume of cork x density of water x g

                        = mass x density of water x g / density of cork

                       = 0.01 x 1000 x 10 / 250 = 0.4 N

Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N

Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N

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3 0
3 years ago
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20 points!
MariettaO [177]
Do they give answer choices? or is it free write? i’ll help if you tell me!!
8 0
3 years ago
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and
frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

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By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

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ε = 23.52

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3 years ago
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