A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but the plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

Answers

Answer 1
Answer:

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

(V_1)/(T_1)=(V_2)/(T_2)\n\Rightarrow {V_2}=(V_1)/(T_1)* T_2\n\Rightarrow {V_2}=(12)/(298.15)* 373.15\n\Rightarrow {V_2}=15.01 L

∴ Final volume at 100°C is 15.01 Liters.


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If a pressure gauge measure an increase in 3×10^(5)Pa on an area of 0.7 m^2, calculate the increase in the force applied to the area?​

Answers

Answer:210000N

Explanation:

Pressure=3x10^5pa

area=0.7m^2

Force = pressure x area

Force=3x10^5x0.7

Force=210000N

How does an increase in cold working effect Modulus of Elasticity and why?

Answers

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

A sound wave travels with a velocity of 330 m/s and has a frequency of 500 Hz. What is itswavelength?

Answers

Answer:

Wavelength = 0.66 meters

Explanation:

Given the following data;

Speed = 330 m/s

Frequency = 500 Hz

To find the wavelength;

Mathematically, wavelength is calculated using this formula;

Wavelength = \frac {speed}{frequency}

Substituting into the equation, we have;

Wavelength = \frac {330}{500}

Wavelength = 0.66 meters

What is the density of the paint if the mass of a tin containing 5000 cm3 paint is 7 kg. If the mass of the empty tin, including the lid is 0.5 kg.​

Answers

We are given:

Mass of the Paint bucket (with paint) = 7000 grams

Mass of the paint bucket (without paint) = 500 grams

Volume of Paint in the Bucket = 5000 cm³

Mass of Paint in the Bucket:

To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)

Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)

Mass of Paint = 7000 - 500

Mass of Paint = 6500 grams

Density of the Paint:

We know that density = Mass / Volume

Density of Paint  = Mass of Paint / Volume occupied by Paint

Density of Paint = 6500/5000

Density of Paint = 1.3 grams / cm³

A solid cylinder of cortical bone has a length of 500mm, diameter of 2cm and a Young’s Modulus of 17.4GPa. Determine the spring constant ‘k’. Please explain.

Answers

Answer:

The spring constant k is1.115* 10^(9) N/m

Solution:

As per the question:

Length of the solid cylinder, L = 500 mm = 500* 10^(- 3) = 0.5 m

Diameter pf the cylinder, D = 2 cm = 0.02 m

As the radius is half the diameter,

Radius, R = 1 cm = 0.01 m

Young's Modulus, E = 17.4 GPa = 17.4* 10^(9) Pa

Now,

The relation between spring constant, k and Young's modulus:

kL = EA

where

A = Area

Area of solid cylinder, A = 2\piR(L + R)

0.5k = 17.4* 10^(9)* 2\piR(L + R)

k = (17.4* 10^(9)* 2\pi* 0.01(0.01 + 0.5))/(0.5)

k = 1.115* 10^(9) N/m

Young's modulus, E is the ratio of stress and strain

And

Stress = (Force or thrust)/(Area)

Strain = (length, L)/(elongated or change in length, \Delta L)

Also

Force on a spring is - kL

Therefore, we utilized these relations in calculating the spring constant.