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Shtirlitz [24]
3 years ago
9

COULD SOMEONE PLS DO THESE FOR ME I’LL LOVE U FOREVER AND EVER

Physics
1 answer:
sladkih [1.3K]3 years ago
7 0
I can help you do them for yourself :)
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Why is it important to understand your strength and weakness<br><br>​
S_A_V [24]

Answer: be a better person

Explanation:

6 0
3 years ago
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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)
UkoKoshka [18]

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

  • The single valance electron leads to weak bonds amongst the element's atoms which makes them soft
  • The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

  • Lithium
  • Sodium
  • Potassium etc

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

<em>Find out more at brainly.com/question/18722874.</em>

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an 269 kg object is moved a distance of 1.9 m by a force if 580 j of work is done on the object what is the object acceleration
IrinaK [193]

Answer:

Explanation:

w=f*d=580*1.5=870J

7 0
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I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway
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