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Orlov [11]
3 years ago
13

Which of the following statements are examples of positive peer pressure?

Physics
2 answers:
Crazy boy [7]3 years ago
5 0

Answer:

C and D

Explanation:

have a great day

Lyrx [107]3 years ago
5 0
C and D
Hope this helps!
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Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial
Maru [420]

Our data are,

State 1:

P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K

State 2:

P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3

We know as well that 3BTU=3.16kJ/K

To find the mass we apply the ideal gas formula, which is given by

P_1V_1=mRT_1

Re-arrange for m,

m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Replacing,

T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F

For conservative energy we have, (Cv = 0.718)

W = m C_v = 0.718  \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU

3 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.
weeeeeb [17]

Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

distance moved by the coaster, d = 225 ft = 68.58 m

final velocity of the coaster, v = 80 mi/h = 35.76 m/s

The time taken for the coaster to drop down the hill is calculated as;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\ \times \ 68.58}{9.8} } \\\\t = 3.74 \ s

The work done by friction on the car is calculated as;

W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76  }{3.74} \times \ 68.58\\\\W = 524,582 \ J

Therefore, the work done by friction on the car is 524,582 J.

6 0
3 years ago
Gaining neutrons makes an element?
Kobotan [32]

Answer:

C. Have no change in electrical charge

Explanation:

If a element gains neutron it become an Isotope. The electrical charge do not change for this, only the atomic mass changes when an element gains neutrons.

The electrical charge is affected when there is a variation in the number of electrons or protons in the element.

6 0
3 years ago
HELP ME PLEASE IM DESPERATE Many inland areas around Florida contain extinct marine fossils.
chubhunter [2.5K]

Answer:b

Explanation:

thats what my teacher said

8 0
2 years ago
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