A bag of potato chips contains 2.00 L of air when it is sealed at sea level at a pressure of 1.00 atm and a temperature of 20.0°C. What will be the volume of the air in the bag if you take it with you, still sealed, to the mountains where the temperature is 7.00°C and atmospheric pressure is 70.0 kPa

Answers

Answer 1
Answer:

The volume of the air in the bag of potato chips to the mountains  which is still sealed, 2.766 liters.

What is the gas law?

The gas law is used to show the relationship between the pressure and the temperature of the gases. It can be given as,

PV=nrT

Here, (n) and (r) are the constant. Therefore,

(PV)/(T)=\rm Constant

For the initial and final values, the gas law can be given as,

(P_1V_1)/(T_1)=(P_2V_2)/(T_2)

Here, (subscript 1,and 2) is used for the initial and final amount of pressure and temperature.

The initial values of the bag of potato chips as volume of 2.00 L, pressure of 1.00 ATM and a temperature of 20.0°C.  It is known that the value of 1 ATM is equal to the 101.325 kPa.

The final temperature of the pack is 7.00°C and atmospheric pressure is 70.0 kPa

Put the values in the above formula as,

(101.325*2)/(293)=(70* V_2)/(280)\nV_2=2.766\rm \; liters

Hence, the volume of the air in the bag of potato chips to the mountains  which is still sealed, 2.766 liters.

Learn more about the gas law here;

brainly.com/question/25290815

Answer 2
Answer:

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume V_(1) = 2.00\ L

Pressure P_(1)= 1.00\ atm

Pressure P_(2)= 70.0\ kPa

Temperature T_(1)= 20.0°C = 293\ K

Temperature T_(2)= 7.00°C = 280\ K

We need to calculate the volume at mountains

Using  gas law

(PV)/(T)=\ Constant

For both temperature,

(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))

Put the value into the formula

(101.325*2)/(293)=(70* V_(2))/(280)

V_(2)=(101.325*2*280)/(293*70)

V_(2)=2.766\ litre

Hence, The volume at mountains is 2.766 L.


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What is the angular width of a person's thumb viewed at arm's length? Assume that the width of the thumb is 17.3 mm and that the distance between the eyes and the thumb is 71.9 cm. Use the small-angle approximation and then convert the answer to degrees.

Answers

Answer:

\theta_(degrees) =0.024\°=0\°1'26.62''

Explanation:

To solve the problem it is necessary to take into account the concepts related to arc length and the radius that make up the measurements of an angle.

An angle is given by the length of arc displaced as a function of the radius, that is

\theta = (Arc_(length))/(Radius)

\theta = (17.3*10^(-3))/(71.9*10^(-2))

\theta = 0.02406rad

360° is equal to do 2\pi rad, therefore:

\theta_(degrees) = 0.02406rad *((180\°)/(2\pi rad))

\theta_(degrees) =0.024\°=0\°1'26.62''

A cylinder with a diameter of 2.0 in. and height of 3 in. solidifies in 3 minutes in a sand casting operation. What is the solidification time if the cylinder height is doubled? What is the time if the diameter is doubled?

Answers

Answer:

3 min 55 sec is the solidification time if the cylinder height is doubled

7min 40 sec if the diameter is doubled

Explanation:

see the attachment

A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

Answers

Answer:

19.3 m/s

Explanation:

Take down to be positive.  Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

v = 19.3 m/s

Calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Answers

Answer:

For 25-turn electromagnet, Number of clips = 4.1

For 50-turn electromagnet number of clips = 9.6

Explanation:

To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Hence;

Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get

For 25-turn electromagnet y = 3.663x * 0.5

(rounded to one decimal place) Number of clips = 4.1

For 50-turn electromagnet y = 7.133x 2.5

(rounded to one decimal place) Number of clips = 9.6

A T-junction combines hot and cold water streams ( = 62.4 lbm/ft3 , cp = 1.0 Btu/lbm-R). The temperatures are measured to be T1 = 50 F, T2 = 120 F at the inlets and T3 = 80 F at the exit. The pipe diameters are d1 = d3 = 2" Sch 40 and d2 = 1¼" Sch 40. If the velocity at inlet 1 is 3 ft/s what is the mass flow rate at inlet 2? (3.27 kg/s)?

Answers

Answer:

m2=3.2722lbm/s

Explanation:

Hello!

To solve this problem follow the steps below

1. Find water densities and entlapies  in all states using thermodynamic tables.

note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)

through prior knowledge of two other properties, such as pressure and temperature.

D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3

D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3

D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3

h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm

h2=Enthalpy(Water;T=120;x=0)=88  BTU/lbm

h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm

2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit

m1+m2=m3

3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out

m1h1+m2h2=m3h3

18.05(m1)+88(m2)=48.03(m3)

divide both sides of the equation by 48.03

0.376(m1)+1.832(m2)=m3

4. Subtract the equations obtained in steps 3 and 4

m1            +      m2       =  m3

-

0.376m1   +  1.832(m2) =m3

--------------------------------------------

0.624m1-0.832m2=0

solving for m2

(0.624/0.832)m1=m2

0.75m1=m2

5. Mass flow is the product of density by velocity across the cross-sectional area

m1=(D1)(A)(v1)

internal Diameter for  2" Sch 40=2.067in=0.17225ft

A=(\pi )/(4) D^2=(\pi )/(4) (0.17225)^2=0.0233ft^2

m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s

6.use the equation from step 4 to find the mass flow in 2

0.75m1=m2

0.75(4.3629)=m2

m2=3.2722lbm/s

The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

W=mg

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

m=0.0001 kg is the mass of the honeybee

g=9.8 m/s^2 is the acceleration due to gravity

Substituting into the equation, we find:

W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N

weight = mg 
here m = 0.000100 g = 9.80 
hence weight = 0.00980 kgm/s2