Describe how you could engineer the situation to produce even more friction and heat

A wooden cylinder (0 02 x 0 02 x 0 1m) floats vertically in water with one-third of ts length immersed. a)-Determine the density

Anuta_ua [19.1K]

Answer:

a)- the density of wood is 333.33 Kg/m³

b)-unstable condition

c)-unstable condition

Explanation:

Given data

wooden cylinder = 0.02m x 0.02m x 0.1m

floating = 1/3 × Length

to find out

density of wood, is it stable condition and wood would float stably in alcohol with density 700 kg/m3

Solution

First we find out density of wood

we know density of water is 1000 kg/m³

and we know wood float 1/3 of length so fraction of density will be

density of wood/ density of water = 1/3

density of wood = 1/3 × density of water

density of wood = 1/3 × 1000 = 333.33 Kg/m³

Now in 2nd part we know for stable condition in partially submerged of body the metacentric height is greater than the centroid of body

we check these condition now,

metacentric height (GM)= I/v

I = ( 0.02 × 0.02³ / 12 )

v = ( 0.02 × 0.02 × 0.1 )

(GM)= I/v = ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

and we know centroid of body (BM) = 0.05 - 0.033 = 0.017

we know height is 0.1m so G act at 0.05 and B act at (0.1 × 0.3 ) = 0.033

we can see that now metacentric height is less than centroid of body so our body is unstable condition

Now in 3rd part we use alcohol so we calculate ratio of density of wood and density of alcohol i.e. = 333.33 / 700 = 0.48

so now our new G will be 0.05 and B will be (0.1 × 0.48 ) = 0.048

metacentric height (GM)= I/v = ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

centroid of body (BM) = 0.05 - 0.048 = 0.002

we can see that now metacentric height is less than centroid of body so our body is unstable condition

5 0

3 years ago

Which of the following would be addressed by an employer completing an EAP template?

zloy xaker [14]

I can help you, but what are the options that were given to you?

4 0

2 years ago

You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(

Vsevolod [243]

Answer:

Output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Explanation:

Given:

dc gain $A = 23.6$ dB

Input signal $V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)$

Now convert gain,

$A = 10^{\frac{23.6}{20} } = 15.13$

DC gain at frequency $f = 0$ is given by,

$A = \frac{V_{out} }{V_{in} }$

$V_{out} =AV_{in}$

$V_{out} = 15.13 \times 0.18 \cos (2\pi (57000)t +18.3)$

At zero frequency above equation is written as,

$V_{out} = 2.72 \times \cos 18.3$

$V_{out} = 2.72$

Now we write output voltage as input voltage,

$V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Therefore, output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

7 0

2 years ago

An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out

Elenna [48]

Answer: The energy system related to your question is missing attached below is the energy system.

answer:

a) Work done = Net heat transfer

Q1 - Q2 + Q + W = 0

b) rate of work input ( W ) = 6.88 kW

Explanation:

Assuming CPair = 1.005 KJ/Kg/K

<u>Write the First law balance around the system and rate of work input to the system</u>

First law balance ( thermodynamics ) :

Work done = Net heat transfer

Q1 - Q2 + Q + W = 0 ---- ( 1 )

rate of work input into the system

W = Q2 - Q1 - Q -------- ( 2 )

where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw

Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw

Q = 18 Kw

Insert values into equation 2 above

W = 6.88 Kw

5 0

2 years ago

A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast

insens350 [35]

Answer:

a) The final equilibrium temperature is 83.23°F

b) The entropy production within the system is 1.9 Btu/°R

Explanation:

See attached workings

8 0

3 years ago