A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J

Explanation:

15.556 metres per second

Answer:

one thousand-millionth of a second.

A nanosecond is an SI unit of time equal to one billionth of a second, that is, ​¹⁄₁ ₀₀₀ ₀₀₀ ₀₀₀ of a second, or 10⁻⁹ seconds. The term combines the prefix nano- with the basic unit for one-sixtieth of a minute. A nanosecond is equal to 1000 picoseconds or ​¹⁄₁₀₀₀ microsecond.  

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

Answer to the following question is as follows;

Explanation:

The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.

Answer:

1) attached below

2) assumption that the earth is spherical

Explanation:

1) Four illustrations of a globe

attached below

2) Reason for distortions at areas that do not fall on lines of tangency or secancy

The reason for distortion on areas outside the lines of tangency or secancy is because of the assumption that the earth is spherical which is not true hence map projections on the areas that fall on the lines of tangency do not experience distortion and are true

Answer:

the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

Given;

dimension of the room, = 15 ft by 15 ft by 20 ft

density of air in the room, ρ = 0.0724 lbm/ft³

The volume of air in the room is calculated as;

Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³

The mass of the air is calculated as;

mass = density x volume

mass = 0.0724 lbm/ft³  x  4,500 ft³

mass = 325.8 lb-m

The weight of the air is calculated as;

Weight = mass x gravity

Weight = 325.8 lb-m x 32.174 ft/s²

Weight = 10482.29 lbm.ft/s²

The weight of the air in pound-force (lb-f) is calculated as;

1 lbf = 32.174 lbm.ft/s²

[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]

Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation:

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

Answer:

4 kg of force

Explanation:

Force = (mass x distance to fulcrum) / length of fulcrum to end

Subsitute values

F = (10 x 20)/50

F =4

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Full Question:

What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?

A) 0°F

B) 273 K

C) 0 K

D) 100°C

E) 273°C

Answer:

The correction Option is D) 100°C

Explanation:

The temperature above is referred to as the critical point.

it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.

There is also a condition under which water can exist in its three forms: that is  

- Ice (solid)

- Liquid (fluid)

- Gas (vapor)

That state is called triple point. The conditions necessary for that to occur are:

Cheers

Cheers

Answer:

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Explanation:

The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:

[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)

Where [tex]t[/tex] is the time, measured in seconds.

The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:

[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)

Then we proceed to integrate on the function in time:

[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]

[tex]\Delta \theta = 20\,rad[/tex]

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Relative density is the ratio of the density of a substance to the density of a given material.

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.

It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.

To learn more about force refer to the link:

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Answer:

560.54 Ω

Explanation:

Applying,

V = IR'............... Equation 1

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

R' = V/I............ Equation 2

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

Answer:

Power = Work / Time

P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts

Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Answer:

Option (e)

Explanation:

A body executing SHM moves to and fro or back and forth  about its mean position.

When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.

So, when it is at maximum distance:

a.

The acceleration is maximum.

b.

The potential energy is maximum.

c.

The total mechanical energy is non zero.

d.

The kinetic energy is zero.

e. The speed is zero. Correct

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

F = Kq₁q₂ / r²

Where

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

https://brainly.com/question/506926

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Answer:

i think the answer is 0.001m³

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0

∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

• P = 350 N

∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0

∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

To learn more about friction and its types:

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