# A Car Tire Is Filled With Air At Pressure 325000 Pa At 303 K. When The Tire Cools,its Pressure Reduces (2023)

Physics High School

new temperature of the tire will be 278.76 K

Explanation:

when the temperature increases, the particles will have greater kinetic energy and also the pressure will be increase for the gas particles.

so when the temperature increases, pressure will also increase and vice versa

T is directly proportional to P

T1 = initial temperature= 303 k

P1= Initial pressure = 325000 pa

T2= Final temperature= ?

P2= Final pressure = 299000 pa

mathematically

P1/T1= P2/T2

T2= P2 x T1/P1

T2 = 299000 x 303/ 325000= 278.76 k

278.76 K

Explanation:

When the tire is filled:

Pressure = P1= 325000 Pa

Temperature = T1= 303 K

When the tire cools:

P2 = 299000 Pa

T2 = ?

Solution:

Pressure is directly proportional to absolute temperature. Using pressure temperature relation:

P1/T1 = P2/T2

T2 = P2 x T1/P1

= 299000 x (303/325000)

T2= 278.76 K

## Related Questions

out of the two spheres of equal mass one rolls down a smooth inclined plane while other is freely falling both from same height . In which case is the work done more?WHY?

In case of sphere rolls down a smooth inclined plane , work is done by gravity only as in free fall but as it is inclined

So, there comes a factor sin( angle of inclination)

So acceleration becomes g sin( angle of inclination)

Now , As in free fall acceleration is g

And max value of Sin is 1

So ,range of acceleration in smooth inclined plane varies from 0 to g

that is max acceleration or max work done as Work = m × acceleration × displacement

= mg sin ( angle) × displacement

As sin( angle) = height/displacement

So it becomes mg× height

In free fall, Work done = mg × height

So, Free fall work done is same as work done in smooth inclined plane

So, Work done in free fall = work done in smooth inclined plane

The gas in a balloon has P=100000 pa and v=0.0279 m^3. if the pressure increases to 120000 pa at constant temperature, what is the new volume of the balloon?( hint:N and T are constant.) also (Unit=m^3)

0.02325 m^3

Explanation:

P1 = 100000Pa , v1 = 0.0279m^3

P2= 120000Pa . V2 = ?

as mass and temperature are constant using ratio method for Boyle'slaw

P2V2= P1V1

or V2= P1V1/P2

V2 = 100000×0.0279/120000

V2= 0.02325 m^3

a car tire is filled with air at pressure 325000 pa at 283 k. if the tire warms up to 302 k, what is the new pressure of the tire (unit = pa)

346819 Pa or ,347000 Pa in 3 significant figures

Explanation:

P1= 325000Pa , T1= 283K ,

P2=? T1= 302 K

as here volume and mass both are constant so using ratio method for pressure temperature law we have P1/T1 = P2/T2

THIS WE CAN ALSO OBTAIN BY RATIO METHOD FOR GENERAL GAS LAW AS

P1V1/(m1T1 ) = P2 V2/ (m2 T2)

IF V1 = V2 =V AND m1=m2=m then expression reduces to

P1/T1 = P2/T2

or P2 = (P1/T1)×T2

P2 = (325000/283) × 302

P2 = 1148.41×302

P2=346819

P2 = 347000 Pa in 3 significant figures

347000

Explanation:

I just took an exam with this question on it and found this to be the correct answer. For how to solve it look at the above answer.

4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box? Make sure to put P,V,and T in the right units)

(Unit=m^3)

Volume of the container = 0.0618 m³

Explanation:

Assuming that the gas is ideal, we can apply ideal gas equation:

PV=nRT

where P = pressure of the gas

V = volume occupied by the gas

n = number of moles of gas

R = Universal gas constant = 8.314 J/mol-K

T = Temperature of the gas

Here we have to find volume.

Given: P=2.25 atm T=385 K n=4.39 moles

Putting the above values in the ideal gas equation, we get:

Volume of the gas = volume of the container in which it is filled = 0.0618 m³

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V = 0.0619 m^3

how many particles are there in 0.622 moles of gas. (answer = _______10^23. just fill in the blank, dont worry about the power.)

1 mole has 6.023× 10^23 particles

So, In 0.622 mole no of particles = 6.023× 0.622× 10^23

=3.74 × 10^23

Explanation:

A scientific law is ____________________A. A rule enacted by the National Academy of Sciences. B. What scientists expect will always happen under particular circumstances. C. A form of scientific inquiry. D. A well-accepted scientific theory.

Your answer would be B). What scientists expect will always happen under particular circumstances

A scientific law is something that will always happen, it is mostly like a repeat when doing a experiment with certain factors in place. This is also known as a "natural law" because something will naturally happen when it is enacted with the same factors for multiple "trials". This "scientific law" will allow scientist to predict what would happen during other multiple trials that are conducted. Concise data would be received from these trials.

An example of a scientific law is Newton's law of gravitation.

Table sugar has a chemical composition of c12h22o11. what is the mass of 8.88 moles of sugar (unit=kg)

3.04 kg up to 3 significant figures

Explanation:

C12H22O11 = 12(12)+22(1)+11(16)= 342 grams = 342÷1000= 0.342 kg

AS atomic mass of carbon is 12 , hydrogen is 1 and of oxygen is 16

now 1 mole has a mass of 0.342 kg

8.88 moles have mass 0.342×8.88=3.04 kg up to 3 significant figures

Explanation:

The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

2.87

Explanation:

What is the molecular mass of ozone o3?

(unit = kg/mol)

To Calculate Molecular Mass of O3

you should know Atomic mass of O

Atomic mass of oxygen(O) = 16

As O3 consist 3 atoms ,

So, Molecular mass of O3 =>3× 16

=> 48 g/mol

=> 0.048 kg/mol

A copper rod is 0.450 m long. Heat is added to the rod until it expands by 0.00130 m. How much did the temperature change? (Unit=C)

170⁰C

Explanation:

α= ΔL/(lo×Δt)

arranging for Δt

Δt =ΔL/(lo×α)

here Δt is change in temperature, ΔL is change in length,lo is original length and is α coefficient of linear expansion and for copper its value is 17×10^-6

Δt =(0.00130)/(0.450×17×10^-6)

Δt=169.9⁰C

Δt= 170⁰C

An aluminum rod and a nickel rod are both 5.00 m long at 20.0 degree Celsius. The temperature of each is raised to 70.0 degrees Celsius. What is the difference in length between the two rods (Unit=m)?

The difference in length between the two rods is 0.0025m.

What is the coefficient of linear expansion?

It defines the material's property of expanding when temperature is increased.

The change in length related to the coefficient of linear expansion is given by

ΔL = αLΔT

where ΔT is the change in temperature, L is the original length and ΔL is the change in length.

An aluminum rod and a nickel rod are both 5.00 m long at 20.0°C. The temperature of each is raised to 70.0°C

α aluminum = 23*10^-6

α nickel = 13*10-6

ΔT = 70.0-20.0 = 50.0°C

L=5.00 m

Substitute the values, we get the change in length as

For Aluminum rod :

ΔL=23*10^-6*5.00*50.0 = 0.00575 m

For Nickel rod :

ΔL=13*10^-6*5.00*50.0 = 0.00325 m

The difference in length between the two rods will be

0.00575m -0.00325m = 0.0025m

Thus, difference in length between the two rods is 0.0025m.

brainly.com/question/14780533

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0.0025 m

Explanation:

If you use the equation ΔL = αLΔT, where ΔL equals the change in length, α equals the linear expansion coefficient, L equals the original length of the object, and ΔT equals the change in temperature, we can calculate the difference in length between the two rods.

α of aluminum = 23*10^-6

α of nickel = 13*10-6

ΔT = 50.0 degrees (derived from the 70.0-20.0)

L=5.00

Once you have gathered all of this given/known information, you may substitute it into the equation. The equation is set up to solve for the change in length so we do not need to do any rearranging. We need to solve one at a time first!

Aluminum:

ΔL=23*10^-6*5.00*50.0 = 0.00575

Nickel:

ΔL=13*10^-6*5.00*50.0 = 0.00325

The final step for solving the difference is simply subtracting the two values we just got to find the difference (similar to what we did when solving for the difference in T):

0.00575-0.00325 = 0.0025

And that's it!! Hope this was helpful!!!

A copper rod is 0.450 m long. Heat is added to the rod until it expands by 0.00130 m. How much did the temperature change? (Unit=C)

170 ⁰C

Explanation:

α= ΔL/(LO×ΔT)

HERE α is coefficient of linear expansion which is constant for different materials , its value for copper is 17×10^-6( ⁰C)^-1

ΔL is change in length , Lo is original length , ΔT is change in temperature

re arranging the equation for ΔT we get

ΔT = ΔL/(LO×α)

= (0.00130)/(0.450× 17×10^-6)

= 169.9⁰C

= 170 ⁰C

A basketball goal has a metal pole that is 3.50 m high at 21°C. On a cool day, at -11.5°C, it contracts by 0.00174 m.? What is its coefficient of linear expansion?

Answer is _____ * 10^-5 C^-1. Fill in the number in the blank.

1.53

Explanation:

α=Δl/(l Δt) here α is coefficient of linear expansion Δl is change in length l is original length andΔt is change in temperature

Δl=0.00174m , l=3.50m ,Δt=21-(-11-5)= 21+11.5 =32.5 ⁰C

α = 0.00174 / (3.50×32.5)

= 1.53×10^-5 ( ⁰C)^-1

1.53×10^-5 ( ⁰C)^-1

A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground? ​

60000N

Explanation:

acceleration is change in velocity

a =(v-u)/t where a is acceleration u is initial velocity and v is final velocity

a = (0-60)/5 = -60/5= - 12m/s^2

here minus sign shows that body is decelerating and force is force of friction Now f = ma here f is force of friction m is mass and a is acceleration

f= 5000×- 12= -60000N

MINUS SIGN HERE SHOWS FORCE OF FRICTION

Hence force of friction is 60000N

how much force is needed to lift a 300 kg block up at 2 m/s^2? be careful there are two forces acting on the block.​

Total force is 3600N

Explanation:

The forces acting on the body are: the force causing upward motion and the downward force due to the weight of the body.

Upward force F=ma

m=300kg, a=2m/s2

F=300*2

=600N

Downward force F=mg

m=300kg, g=10m/s2

F=300*10

=3000N

The total force needed to lift the load is:

3000+600=3600N

3543N OR 3540 N IN 3 SIGNIFICANT NUMBERS

Explanation:

Let F be the force which lifts body upward

in downward direction weight of the body acts .

weight = mass ×acceleration due to gravity

weight, W = 300×9.81=2943N

Now net force acting on body is F-W and according to second law of motion

F-W = m a where F is force W is weight m is mass and a is acceleration

F-2943=300×2

F = 600+2943

F = 3543N or 3540 in 3 significant figures

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and mass M-100 kg. In operation, the wheel rotates at 60.0 rev/min. While the wheel is spinning, your grandmother works clay at the center of the wheel with her hands into a pot-shaped object with circular symmetry. When the correct shape is reached, she wants to stop the wheel in as short a time interval as possible, so that the shape of the pot is not further distorted by the rotation. She pushes continuously with a wet rag as hard as she can radially inward on the edge of the wheel and the wheel stops in 6.00 s (a) You would like to build a brake to stop the wheel in a shorter time interval, but you must determine the coefficient of friction between the rag and the wheel in order to design a better system. You determine that the maximum pressing force your grandmother can sustain for 6.00 s is 50.0N. k0.544

(b) What If? If your grandmother instead chooses to press down on the upper surface of the wheel a distance r 0.250 m from the axis of rotation, what is the force (in N) needed to stop the wheel in 6.00 s? Assume that the coefficient of kinetic friction between the wet rag and the wheel remains the same as before (Enter the magnitude.) 25.99 Remember that a torque is a product of a force and a distance. N

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

= Final angular velocity

= Initial angular velocity

= Angular acceleration

Mass of inertia is given by

Angular acceleration is given by

Equation of rotational motion

The coefficient of friction is 0.54454

At r = 0.25 m

The force needed to stop the wheel is 104.00902 N

A child has a toy tied to the end of a string and whirls the toy at constant speed in a horizontal circular path of radius R. The toy completes each revolution of its motion in a time period T. What is the magnitude of the acceleration of the toy?a. g b. 2πg
c. Zero
d. 4π²R/T²
e. πR/T²

d.

Explanation:

= Radius of the circular path

= Time period of motion

= Angular speed of the motion

Angular speed of the motion is given as

The toy is moving in a circle and hence centripetal acceleration acts on it which can be given as

hence the correct choice is

d.

. Gravitational force of attraction "F" exists between two-point masses A and B when a fixed distance separates them. After mass A is doubled and mass B is quartered, the gravitational attraction between the two masses is___________.

Explanation:

Given

Gravitation attraction is F between A and B

if and is the mass of A and B respectively then

F is given by

where r=distance between them

if is doubled and is quadrupled then Force can be given by

thus force becomes 8 times of initial value

0.5F OR HALF OF PREVIOUS VALUE

Explanation:

F= G Ma Mb/r²

now new force F1 is given as

F1= G ×2Ma × 0.25 Mb /r²

F1 = 2×0.25 ×Ma×Mb/r²

F1 = 0.5×Ma Mb/r²

F1= 0.5 F

You are driving a 4000 kg pickup truck on a cold winter day when the coefficient of static friction for tires on the icy road is only 0.13. What is fastest you can drive around a bend whose radius of curvature is 50 m?

8 m/s

Explanation:

= mass of the pickup truck = 4000 kg

= Coefficient of static friction between tires and ice road = 0.13

= Radius of curvature = 50 m

The frictional force between tires and ice road provides the necessary centripetal force to move in circle. hence

centripetal force = frictional force

While a roofer is working on a roof that slants at 39.0 degrees above the horizontal, he accidentally nudges his 88.0 N toolbox, causing it to start sliding downward, starting from rest. If it starts 4.90 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 18.0 N ?

V= 6.974 m/s

Explanation:

Component( box) weight acting parallel and down roof 88(sin39.0°)=55.4 N

Force of kinetic friction acting parallel and up roof = 18.0 N

Fnet force acting on tool box acting parallel and down roof

Fnet= 55.4 - 18.0

Fnet=37.4 N

acceleration of tool box down roof

a = 37.4(9.81)/88.0

a= 4.169 m/s²

d = 4.90 m

t = √2d/a

t= √2(4.90)/4.169

t= 1.662 s

V = at

V= 4.169(1.662)

V= 6.974 m/s

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