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4 years ago

______ clouds are formed when air masses rise vertically in the atmosphere.

Physics

2 answers:

DIA [1.3K]4 years ago

7 0

Answer:

a) Cumulus

Explanation:

zhannawk [14.2K]4 years ago

6 0

a) Cumulus is 100% the correct answer

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There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp

saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

pressure is inversely proportional to volume,

P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

P2, V2 represents the second pressure and volume

P2 = (P1 V1) / V2

= (1.75 $\times$ 8.8) / 6.2

P2 = 2.48 atmosphere.

5 0

3 years ago

Which of the following items can enter and exit an isolated system?

Mkey [24]

Neither energy nor matter can enter and exit an isolated system.

Explanation:

There are three types of systems which refers to universe. They are

1. Open System : In an open system, both energy and matter have external interactions. Example:- boiling water in an open pan.

2. Closed system : In a closed system, only energy has interaction with surroundings. Example:- boiling water with a lid on the pan.

3. Isolated system : In an isolated system, neither energy-nor matter has external interactions with surroundings. Example : a thermos flask does not exchange energy and matter.

Hence, the correct option is (a) " neither energy nor matter ".

7 0

4 years ago

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an

Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

$\delta = \frac{Pl}{AE}$

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

$\delta = \frac{Pl}{AE}$

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$ Poission's ratio

$l_{a}$ = Lateral strain

$l_{s}$ = Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$

5 0

3 years ago

A room has a volume of 160 m3. An air-conditioning system is to replace the air in this room every 21.3 minutes, using ducts tha

serious [3.7K]

Answer:

A. 0.204203928 m

b. 0.1443939822

Explanation:

Givens:

Volume (V) = 160 m^3

velocity 1 (v1) = 3.00 m/s

velocity 2 (v2) = 6.00 m/s

Time (t) = 21.3 minutes = 1278 seconds

Side length = ?

V = A*v*t

A = s^2

V = s^2*v*t

s= $\sqrt{\frac{V}{v*t} }$

s = $\sqrt{\frac{V}{v*t} }$ = $\sqrt{\frac{160}{3.00*1278} }$ = 0.204203928 m

s = $\sqrt{\frac{V}{v*t} }$ = $\sqrt{\frac{160}{6.00*1278} }$ = 0.1443939822 m

I hope this helped!! Have a good day.

7 0

3 years ago

Helpppp please I'm really stuck on this question

amm1812

Is this a test? If it is, i’m not allowed to help as i would be seen as a bad person! I hope you pass, god bless.

6 0

3 years ago

Read 2 more answers