When laser light of wavelength 632.8 nm passes through a diffraction grating, the first bright spots occur at ± 17.8 ∘ from the

Thrust is quantified as a reaction force, in accordance with Newton's third law. When a system accelerates or expels mass in one direction, this resulting mass generates a force of equal strength but in the opposite direction on that system. This relationship can be expressed mathematically as:

Where:

v = velocity of the exhaust gases as perceived from the rocket.

= Change in mass over time

The provided data is as follows:

After substitution, we obtain:

Answer:

Explanation:

Utilizing the formula for surface radiation P (energy per second in Watts) = emissivity constant * surface area * Stefan-Boltzmann constant * Temperature in Kelvin^4 *

2.7*10^31 = 1* 5.67*10^-8*A*11000^4

Rearranging to solve for A = 2.7*10^31 / (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Assuming the shape is spherical, the surface area is = 4πR^2 (radius of Rigel)

R = √(0.3261*10^23 / 4*π) = 5.1 * 10^10m

B) repeating the same calculation

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of Procyon

Rearranging gives A = 2.1*10^23/(5.67*10^-8*10^16)

A = 0.37*10^15

Assuming the star is spherical;

A = 4πR^2 where R is Procyon's radius

R = √(0.37*10^15/4π) = 5.4*10^6m

Answer:

The correct response is:

1. KE Increases, PE Increases, ME Increases.

Explanation:

In this context, kinetic energy refers to the energy associated with an object's motion. Kinetic energy can be defined as the energy required to accelerate a mass from rest to a specified velocity, which it maintains once that speed is reached:

KE = 1/2 mv².

This definition indicates that KE is on the rise.

Potential energy is the energy stored in a body due to its position in a gravitational field:

PE = mgh,

which increases as the object is elevated against gravitational pull.

Since both kinetic and potential energies are increasing, it follows that the total mechanical energy (ME) is also rising:

ME = PE + KE.

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

Substituting the values into the equation

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

Insert the values into the formula

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

Additional explanation

The Ideal Gas Law that should be remembered is:

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

Next, we will determine the work performed on the gas:

Step B:

By utilizing the methodology mentioned earlier:

Next, we will ascertain the work completed on the gas:

Ultimately, we can calculate the total work done and heat supplied as follows:

Learn more

• Minimum Coefficient of Static Friction:
• The Pressure In A Sealed Plastic Container:
• Effect of Earth’s Gravity on Objects:

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure