What does Boyle's Law state about the relationship between the pressure and volume of an ideal gas at constant temperature?
a) The product of pressure and volume increases as pressure decreases.
b) The sum of pressure and volume is constant.
c) The sum of product and volume decreases as volume increases.
d) The product of pressure and volume is constant.

Answers

Answer 1

Answer:

Option (d).

Explanation:

According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.

So,

Let the pressure is P, volume is V and T is the absolute temperature of the gas.

Pressure proportional to the reciprocal of the volume.

[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]

The correct option is (d).


Related Questions

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.

Answers

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

Why is the force of attraction between the Earth and ourselves so huge compared to the attraction between two apples?

Answers

Answer:

Answer in explanation

Explanation:

The force of attraction between two bodies is governed by Newton's Law of Gravitation:

[tex]F = \frac{Gm_1m_2}{r^2}[/tex]

where,

G = Universal Gravitational Constant

m₁ = mass of the first body

m₂ = mass of the second body

r = distance between the two bodies

F = Force

Hence, it is clear from the formula that the magnitude of the force is directly proportional to the product of the masses of the objects. So in the case of the earth and ourselves, the mass of the earth is very large in order of 10²⁴ kg. Due to this huge mass, the attraction between the earth and ourselves is so huge as compared to the attraction between two apples. Because the masses of the apple are very small in grams.

In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?

Answers

Answer:

P = 0.14 hp

Explanation:

The power required by the ship is given as:

[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]

where,

P = Power = ?

m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 2 m

t = time = 1 s

Therefore,

[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]

Converting to horsepower (hp):

[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]

P = 0.14 hp

65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight of the body in air.​

Answers

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s. Calculate
(a) the angular acceleration,
(b) the time required to complete the 60 revolutions,
(c) the time required to reach the 10 rev/s angular speed, and
(d) the number of revolutions from rest until the time the disk reaches the 10 rev/s angular speed.

Answers

Explanation:

Given:

[tex]\omega_0[/tex] = 10 rev/s = [tex]20\pi\:\text{rad/s}[/tex]

[tex]\omega[/tex] = 15 rev/s = [tex]30\pi\:\text{rad/s}[/tex]

[tex]\theta[/tex] = 60 rev = [tex]120\pi\:\text{rads}[/tex]

a) the angular acceleration [tex]\alpha[/tex] is given by

[tex]\alpha = \dfrac{\omega^2 - \omega_0^2}{2\theta}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{(30\pi)^2 - (20\pi)^2}{240\pi} = 6.5\:\text{rad/s}^2[/tex]

b) [tex]t = \dfrac{\omega - \omega_0}{\alpha} = \dfrac{30\pi - 20\pi}{6.5} = 4.8\:\text{s}[/tex]

c) [tex]t = \dfrac{\omega - \omega_0}{\alpha}[/tex]

[tex]=\dfrac{20\pi - 0}{6.5} = 9.7\:\text{s}[/tex]

d)[tex]\theta = \frac{1}{2}\alpha t^2[/tex]

[tex]\:\:\:\:\:\:\:=\frac{1}{2}(6.5\:\text{rad/s}^2)(9.7\:\text{s})^2 = 305.8\:\text{rad}[/tex]

[tex]\:\:\:\:\:\:\:= 48.7\:\text{revs}[/tex]

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Answers

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer to the 3.0uC charge?​

Workdone

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

W=4.49*10e10 J

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Question 2 of 10
Which of the following statements is true of an isolated system?
A.The system has energy but no matter.
B.Energy separates the matter in the system from outside matter.
C.The matter within the system does not interact with matter outside the system
D.The matter within the system does not interact with other matter in the system

Answers

Answer:

D

Explanation:

because it is system removed from system and it dosent interact with them

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?

Answers

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]

Using formula:

[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]

   [tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]

[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]

Calculating the Work by net force

[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]

The above work is converted into thermal energy.

Now,

[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]

There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their

A temperatures.

B volumes.

C densities.

D masses.

Answers

Answer: masses

Explanation:

Trust me

What is the work done if a Boulder of mass 100 kilogram is rolled 40 meter up slope an angle of 20 degrees assuming the force of friction is negligible​

Answers

Answer:

The work done is 13680.8 J.

Explanation:

The work done can be calculated as follows:

[tex] W = F*d [/tex]              

Where:            

F: is the force                                                        

d: is the displacement = 40 m                                    

The force acting on the boulder is given by:

[tex] F = mgsin(\theta) [/tex]

Where:

m: is the mass = 100 kg

g: is the acceleration due to gravity = 10 m/s²

θ: is the angle = 20°      

Then, the work is:

[tex] W = mgsin(\theta)d = 100 kg*10 m/s^{2}*sin(20)*40 m = 13680.8 J [/tex]

Therefore, the work done is 13680.8 J.  

I hope it helps you!  

the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is

Answers

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².

~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)


a.) Determine the linear regression

b.) At this given angular velocity, what is the rotational kinetic energy?

Answers

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the
acceleration of the block?

Answers

Answer:

F = m g sin theta      force accelerating block

m a = m g sin theta

a = 9.8 sin 24 = 3.99 m/sec^2

how did kepler discoveries contribute to astronomy

Answers

Answer:

They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.

Explanation:

A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frictional force of 2 N, find (a) its acceleration, and (b) its initial velocity.

Answers

B it’s Intail velocity

Light energy is part of a larger form of energy known as __________.

Answers

Light energy is part of a larger form of energy known as electromagnetic energy. Details about electromagnetic energy can be found below.

What is electromagnetic radiation?

Electromagnetic spectrum is the entire range of wavelengths of all known electromagnetic radiations extending from gamma rays through visible light, infrared, and radio waves, to X-rays.

Visible light is the part of the electromagnetic spectrum, between infrared and ultraviolet, that is visible to the human eye.

Therefore, Light energy is part of a larger form of energy known as electromagnetic energy.

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What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5

Answers

Answer:

3

Explanation:

You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.

The reactant side of the equation also has three moles:

1 mole of CaCO₃ and 2 moles of HCl.

A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done

Answers

Answer:

W =  343.2 J

Explanation:

Given that,

Mass of bale of hay = 26 kg

Horizontal force exerted = 88 N

Distance moved, d = 3.9 m

Work done, W = Fd

Put all the values,

W = 88 N × 3.9 m

= 343.2 J

So, the work done is 343.2 J.  

A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?

Answers

Answer:

7.5 kg

Explanation:

We are given that

[tex]m_1=10 kg[/tex]

Length of plank, l=3 m

Distance of fulcrum from one end of the plank=1 m

[tex]m_2=20 kg[/tex]

We have to find the mass must be on the other end if the plank remains balanced.

Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]

[tex]20=10(0.5)+2m[/tex]

[tex]20=5+2m[/tex]

[tex]2m=20-5=15[/tex]

[tex]\implies m=\frac{15}{2}[/tex]

[tex]m=7.5 kg[/tex]

Hence, mass 7.5 kg   must be on the other end if the plank remains balanced.

Answer:

The mass at the other end is 7.5 kg.

Explanation:

Let the mass is m.

Take the moments about the fulcrum.

20 x 1 = 10 x 0.5 + m x 2

20 = 5 + 2 m

2 m = 15

m = 7.5 kg

NEED HELP ASAP- Please show work

The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:

(a) (5 points) The angular velocity at t = 2 s?

(b) (5 points) The angular acceleration at t = 2 s?

Answers

Answer:

Look at work

Explanation:

Θ= 4t^3+10t-40

a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.

Θ= 4*8+20-40=12

use ω=Θ/t

Plug in values

ω=6 rad/s

b) In order to find α we use ω/t.

Plug in values

α=6/2= 3 rad/s^2

A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis​

Answers

Answer: -5×10-3

Explanation:

E=kq/r

Wood is an example of
A. Metalloid
B. Insulator
C. Nonmetal
D. Conductor

Answers

The Answer is b: insulator

How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination

Answers

Answer: hello tables and data related to your question is missing attached below are the missing data

answer:

a) I = I₁ = I₂ = I₃ = 0.484 mA

b) I₁ =  0.016 amps

   I₂ =  0.0016 amps

   I₃ = 7.27 * 10^-4 amps

c)  I₁ = 1.43 * 10^-3 amp

    I₂ =  0.65 * 10^-3 amps

Explanation:

A) magnitude of current for Part 1

Resistors are connected in series

Req = r1 + r2 + r3

       = 3300 Ω  ( value gotten from table 1 ) ,

          V = 1.6 V ( value gotten from table )

hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA

The magnitude of current is the same in the circuit

Vi = I * Ri

B) magnitude of current for part 2

Resistors are connected in parallel

V = 1.6 volts

Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) +  R3 ]

      = [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]

      = 87.30 Ω

For a parallel circuit the current flow through each resistor is different

hence the magnitude of the currents are

I₁ = V / R1 = 1.6 / 100 = 0.016 amps

I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps

I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps

C) magnitude of current for part 3

Resistors are connected in combination

V = 1.6 volts

Req = R1 + ( R2 * R3 / R2 + R3 )

       = 766.66 Ω

Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps

magnitude of currents

I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps

I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps

The mass of the moon is 7.2 × 10^22 kg and its radius is 1.7×10^6 m.What will be the gravity of the moon to a body of the mass 1 kg on the surface of the moon.​

Answers

Answer:

1.66 N

Explanation:

The force of gravity of the moon on the body is given by

F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.2 × 10²² kg, m = mass of body = 1 kg and R = radius of moon = 1.7 × 10⁶ m

Substituting the values of the variables into the equation, we have

F = GMm/R²

F = 6.67 × 10⁻¹¹ Nm²/kg² × 7.2 × 10²² kg × 1 kg/(1.7 × 10⁶ m)²

F = 48.024 × 10¹¹ Nm²/2.89 × 10¹² m²

F = 16.62 × 10⁻¹ N

F = 1.662 N

F ≅ 1.66 N

So, the gravity on the moon is 1.66 N

The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.

Answers

Answer:

[tex]\frac{r_1}{r_2}=6.9[/tex]

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]

I need help with this physics question.

Answers

The acceleration will increase by 61.3%.

Explanation:

The centripetal acceleration [tex]a_c[/tex] is given by

[tex]a_c = \dfrac{v^2}{r}[/tex]

If the velocity of the object increases by 27.0%, then its new velocity v' becomes

[tex]v' = 1.270v[/tex]

The new centripetal acceleration [tex]a'_c[/tex] becomes

[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]

[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]

what are the dynamic properties of a nucleus​

Answers

Dynamic properties of nucleus are:
It contains protons and neutrons .
It’s still stable even though it has only positive charge because of specific binding energy.
There exists nuclear forces in the nucleus.
It attracts and holds electrons around it.
It’s of size around 10 power -15 in metres.
It is very denser space in an atom. Etc

A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?

Answers

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

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The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?
a) shows a lot of negative signs in the interior of circle
b) shows a lot of positive signs in the interior of circle
c) shows a hollowed out "hole" in the interior of the circle, with negative signs surrounding the opening.
d) shows a hollowed out "hole" in the interior of the circle, with positive signs surrounding the exterior edge

Answers

Answer:

d

Explanation:

The minimum energy configuration in electrostatics states that Charges always reside on the surface of a conductor. If anyhow they were inside, an electric field would exist inside and would act to move them to the surface,  

Therefore, the drawing that shows where the charges reside when they are in equilibrium is a hollowed-out "hole" in the interior of the circle, with positive signs surrounding the exterior edge. This means that the d part is the correct answer.

boat carrying people more than its capacity is attributes of sinking why​

Answers

It will sink because of how much the max capacity a boat can have it will sink because it is over capacity that’s why it will sink.

Hope this helps :)

Answer:

Upthrust on boat becomes lesser than Weight of boat

Explanation:

When there are more people than the capacity, The weight of the boat acting downwards increases. However, the upthrust acting on the submerged part of the boat is constant. Since Weight > Upthrust, there is a net force downwards, leading to sinking.

Other Questions
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