How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?Round off the answer to the nearest whole number.

Answers

Answer 1

Answer:

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=(0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m)/((\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m) \n\ny=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=(1)/(12)*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\n0.888)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+(1)/(12)*3(1.5)(1.5)^2+\n3(1.5)(0.888-0.75)^2\n\nI_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=(1)/(12)*3(0.8)(0.8)^2+ (1)/(3)* 3(1.5)(1.5)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\n(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\n\nI_o=13.4\ kgm^2$

Final answer:

To solve this problem, calculate the mass of each element of the pendulum, use that information to determine the center of mass, and then apply the parallel axis theorem to calculate the two moments of inertia.

Explanation:

To determine the center of mass and the mass moment of inertia of the pendulum, first we calculate the individual masses of the rods: AB and OC, and the plate. Each rod has a mass of 2 kg (given mass per unit length is 3kg/m and length of each rod is 1 m from the first reference paragraph).

The center of mass ӯ can be determined using the formula for center of mass, averaging distances to each mass element weighted by their individual masses. The mass moment of inertia, also known as the angular mass, for rotation about the z axis through G is determined using the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance D away from an axis through the center of mass is the sum of the moment of inertia for rotation about the center of mass and the total mass of the body times D squared.

Finally, the moment of inertia about the z axis passing through the center of rotation O can be calculated again using the parallel axis theorem, with distance d being the distance between points G and O.

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A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the beam make with the normal in the glass?

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To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

$n_1sin\theta_1 = n_2 sin\theta_2$

Where,

n_i = Refractive indices of each material

$\theta_1$ = Angle of incidence

$\theta_2$ = Refraction angle

Our values are given as,

$\theta_1 = 38\°$

$n_1 = 1$

$n_2 = 1.4$

Replacing

$1*sin38 = 1.4*sin\theta_2$

Re-arrange to find $\theta_2$

$\theta_2 = sin^(-1) (sin38)/(1.4)$

$\theta_2 = 26.088°$

Therefore the angle will the beam make with the normal in the glass is 26°

One of the primary visible emissions from a distant planet occurs at 425 nm. Calculate the energy of a mole of photons of this emission.]

Answers

Answer:

Explanation:

Given

Wavelength of incoming light $\lambda =425\ nm$

We know

$speed\ of\ wave=frequency* wavelength$

$frequency=(speed)/(wavelength)$

$\mu =(3* 10^8)/(425* 10^(-9))$

$\mu =7.058* 10^(14)\ Hz$

Energy associated with this frequency

$E=h\mu$

where h=Planck's constant

$E=6.626* 10^(-34)* 7.058* 10^(14)$

$E=46.76* 10^(-20)\ Hz$

Energy of one mole of Photon $=N_a* E$

$=6.022* 10^(23)* 46.76* 10^(-20)$

$=281.58* 10^(3)$

$=281.58\ kJ$

Final answer:

To calculate the energy of a mole of photons of the emission at 425 nm, use the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. Convert the wavelength to meters, substitute the values into the equation, and calculate to find the energy of a single photon. Multiply this by Avogadro's number to find the energy of a mole of photons.

Explanation:

To calculate the energy of a mole of photons of the emission at 425 nm, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (in meters).

Converting the wavelength to meters, we have 425 nm = 425 x 10^-9 m.

Substituting the values into the equation, we get E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m). Calculating this gives us the energy of a single photon of this emission. To find the energy of a mole of photons, we can multiply this value by Avogadro's number (6.02 x 10^23 photons/mol).

Therefore, the energy of a mole of photons of this emission is (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (425 x 10^-9 m) x (6.02 x 10^23 photons/mol).

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A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

Answers

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

$F=(k.q.Q)/(L)$

where:

F is the electric force on the point charge,

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The chargedistribution remains the same, only the length changes.

So, the new electric force $F_f$ on the proton after the segment is shrunk becomes:

$F_f=(k.q.Q)/((1)/(3)L)$

The original electric force $F_i$ on the proton before the segment was shrunk is:

$F_i = (k.q.Q)/(L)$

let's find the ratio $(F_f)/(F_i)$ :

$(F_f)/(F_i)=((k.q.Q)/((1)/(3)L))/((k.q.Q)/(L))$

$(F_f)/(F_i)=3$

Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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Final answer:

The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

Explanation:

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

A physicist is creating a computational model of a falling person before and after opening a parachute. What boundary conditions would be important here?the air resistance encountered as the person falls

the speed at which the person falls

the change in kinetic and potential energy

the location where potential energy is zero

Answers

Answer:

the location where potential energy is zero

Explanation:

Answer:

Air resistance

Explanation:

Air resistance encountered as the person falls

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by: