When an ideal gas undergoes a slow isothermal expansion, A : the work done by the environment is the same as the energy absorbed as heat. B : the increase in internal energy is the same as the work done by the environment. C : the work done by the gas is the same as the energy absorbed as heat. D : the increase in internal energy is the same as the heat absorbed. E : the increase in internal energy is the same as the work done by the gas.

Answer:

a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².

Explanation:

a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.

τ = Frsinθ

= 185 N × 1.73 m × sin90°

= 320.05 Nm

So, the torque τ = 320.05 Nm

b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m

α = angular acceleration of grindstone

τ = Iα

α = τ/I = τ/(MR²/2) = 2τ/MR²

substituting the values of the variables, we have

α = 2τ/MR²

= 2 × 320.05 Nm/[76 kg × (1.73 m)²]

= 640.1 Nm/227.4604 kgm²

= 2.814 rad/s²

So, the angular acceleration α = 2.814 rad/s²

c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.

So  τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm

The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm

Since τ'' = Iα

α' = τ''/I where α' = its new angular acceleration

α' = 2τ/MR²

= 2 × 319.75 Nm/[76 kg × (1.73 m)²]

= 639.5 Nm/227.4604 kgm²

= 2.811 rad/s²

So, the angular acceleration α' = 2.811 rad/s²

Answer:

The bumper will be able to move by 0.01155m.

Explanation:

The magnitude of deceleration of the car in the front end collision.

[tex]a = \frac{F_m}{m} \\[/tex]

[tex]a = \frac{80000}{1500} \\[/tex]

[tex]a = 53.33[/tex]

This is the deceleration of the car that is generated to stop due to a front end collision.

4 km/h = 1.11 m/s

Now, the initial speed of the bumper in the relation of car, Vi = 0

Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s

Use the below equation:

[tex]s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\[/tex]

[tex]s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\[/tex]

[tex]s = 0.01155 \\[/tex]

Thus, the bumper can move relative to the car is 0.01155 m .

-- static friction

-- kinetic friction

-- fluid friction

-- sliding friction

-- air resistance

-- drag

-- professional debate

Answer:

Answer:

A. Increasing the number of lines per length.

Answer:

Explanation:

At the point midway between wires

magnetic field due to wire having current 2I₀

= 10⁻⁷ x 2 x2I₀ / r     where 2r is the distance between wires .

magnetic field due to wire having current I₀

= 10⁻⁷ x 4 I₀ / r

magnetic field due to wire having current I₀

= 10⁻⁷ x 2I₀ / r    

= 10⁻⁷ x 2 I₀ / r     where 2r is the distance between wires .

these fields are in opposite direction as direction of current is same in both .

net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r

= 2 x 10⁻⁷ x  I₀ / r

At point A net magnetic field = 2 x 10⁻⁷ x  I₀ / r

At point B , we shall calculate magnetic field

magnetic field due to nearer wire having current  2 I₀ = 10⁻⁷ x 4 I₀ / r

magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r

These magnetic fields act in the same direction so they will add up

net magnetic field = [ (4 I₀ / r)  + (2 I₀ / 3r) ] x 10⁻⁷

= (14 I₀ / 3r ) x 10⁻⁷

Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷

Ratio of field at A and B

= 3 / 7 . Ans

The ratio of the magnitude of the magnetic field at point A to point B is :

3 / 7

Given data :

Magnitude of the left current is  2I₀

Magnitude of the right current is  I₀

First step : Determine the magnetic field at point A  

The magnetic field due to the left current ( 2I₀ )

10⁻⁷ * 2 * 2I₀ / r       ( 2r = distance between wires )

The magnetic field due to the right current ( I₀ )

10⁻⁷ * 2 I₀ / r

From the expressions above the magnetic fields are in  opposite direction

∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r =   2 * 10⁻⁷ *  I₀ / r

Hence The magnetic field at point A = 2 * 10⁻⁷ *  I₀ / r

Next step : determine the magnetic field at point B

Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r

Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r

Since the fields acts in the same directions

The net magnetic field =  (4 I₀ / r)  + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷

Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷

Therefore the ratio of the magnitude of the magnetic field at point A to point B  =  3/ 7

Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B  = 3 / 7

Learn more : https://brainly.com/question/22403676

Answer:

8.1 m

Explanation:

Convert km/h to m/s.

45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s

Distance = speed × time

d = (12.5 m/s) (0.65 s)

d = 8.125 m

Answer:

Explanation:

a) To get a rainbow from a prism arrangement, we will need

First, you cut a slit in one end of the cardboard with the pen knife.

Next you open up a space on top of the cardboard through which you can observe the experiment and its result.

Next, you place the triangular prism with its slant face facing the the cut slit.

Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.

b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.

Answer:

[tex]density \: = \frac{mass}{volume} [/tex]

1 / 5.587 is equal to 0.179 kg/m³

Hope it helps:)

Answer:

The answer is

Explanation:

Density of a substance is given by

[tex]Density \: = \frac{mass}{volume} [/tex]

From the

mass = 1 kg

volume = 5.583 m³

Substitute the values into the above formula

We have

[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]

We have the final answer as

Hope this helps you

Answer:

65.3

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer:

It will be 》》》》1.664716m

Answer:

Explanation:

A) Vair = 1.3 L

B) Volume is not reasonable

Explanation:

A)

Assume

m to be total mass of the man

mp be the mass of the man that pulled out of the water

m1 be the mass above the water with the empty lung

m2 be the mass above the water with full lung

wp be the weight that the buoyant force opposes as a result of the air.

Va be the volume of air inside man's lungs

Fb be the buoyant force due to the air in the lung

given;

m = 78.5 kg

m1 = 3.2% × 78.5 = 2.5 kg

m2 = 4.85% × 78.5 = 3.8kg

But, mp = m2- m1

mp = 3.8 - 2.5

mp = 1.3kg

So using

Archimedes principle, the relation for formula for buoyant force as;

Fb = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

Fb = wp = 1.3× 9.81

Fb = 12.7N

But

Fb = (ρ_water × V_air × g)

So

Vair = Fb/(ρ_water × × g)

Vair = 12.7/(1000 × 9.81)

V_air = 1.3 × 10^(-3) m³

convert to litres

1 m³ = 1000 L

Thus;

V_air = 1.3× 10^(-3) × 1000

V_air = 1.3 L

But since the average lung capacity of an adult human being is about 6-7litres of air.

Thus, the calculated lung volume is not reasonable

Explanation:

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

Answer:

69.58 m tall

Explanation:

Pls see attached file

Answer:

A)  The distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B)  The electrical force of repulsion among them shows a small effect  in beam stability.

Explanation:

Given that:

A Van de Graaff generator produces a beam of 2.02-MeV deuterons

If the beam current is 10.0 μA, the distance of the deuterons from one another can be determined by using the concept of kinetic energy of the generator.

[tex]\mathtt{K.E = \dfrac{1}{2}mv^2}[/tex]

2 K.E = mv²

[tex]\mathtt{v^2 = \dfrac{2 K.E }{m}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{2 K.E }{m}}}[/tex]

so, v is the velocity of the deuterons showing the distance of the deuterons apart from one another.

[tex]\mathtt{v =\sqrt{ \dfrac{2 (2.02 \ MeV) \times \dfrac{10^6 \ eV}{ 1 \ MeV} \times \dfrac{1.60 \times 10^{-19} \ J }{1 \ eV} }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

[tex]\mathtt{v =\sqrt{ \dfrac{6.464 \times 10^{-13} \ J }{ 3.34 \times 10^ {-27} \ kg}}}[/tex]

v = 13911611.49  m/s

v = 1.39 × 10⁷ m/s

So, If the beam current is 10.0 μA.

We all know that:

[tex]I = \dfrac{q}{t}[/tex]

[tex]t = \dfrac{q}{I}[/tex]

[tex]\mathtt{ t = \dfrac{1.6 * 10 ^{-19} \ C}{10.0 * 10^{-6} \ A}}[/tex]

t = 1.6 × 10⁻¹⁴ s

Finally, the distance of the deuterons from one another  = v × t

the distance of the deuterons from one another  = (1.39 × 10⁷ m/s × 1.6 × 10⁻¹⁴ s)

the distance of the deuterons from one another  = 2.224× 10⁻⁷ m

B) Is the electrical force of repulsion among them a significant factor in beam stability? Explain.

The electrical force of repulsion among them shows a small effect  in beam stability. This is because, one nucleus tends to put its nearest neighbor at potential V = (k.E × q) / r = 7.3e⁻⁰³ V. This is very small compared to the 2.02-MeV accelerating potential, Thus, repulsion within the beam is a small effect.

Explanation:

According to Thermal Expansion of solids:

[tex]dl = \alpha \times l \times dt[/tex]

[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]

[tex]dl = 7.2 \times {10}^{ - 3} [/tex]

Answer:

C $0.75 my friend I wish it is right answer

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

Answer:

A) the frequency and amplitude of the output voltag

Explanation:

Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.

Changing the speed regulator will change the engine throttle setting to maintain the speed.

While the power, torque, current, fuel flow rate and torque angle will have decreased.

1. The ocean water collects back in the ocean.

2. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.

3. an excessive amount of water flowing from downslope along earths surface

4. A.Evaporation occurs when water is warmed by the sun.

5. The water returns into the ocean by the water cycle . It evaporates , then it condensates , then it participates ( Rains ) and then goes back into the ocean.

Hope this answer correct ✌️

Answer:

27yrs

Explanation:

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Potential Energy = 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Answer:

(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV

(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Explanation:

Given;

radius of the circular loop, r = 31.0 cm = 0.31 m

initial magnetic field, B₁ = 0.7 T

final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T

duration of change in the field, t = 29

(a) The magnitude of induced emf in the loop while the magnetic field is increasing.

[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]

[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]

Where;

A is the area of the circular loop

A = πr²

A = π(0.31)² = 0.302 m²

[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]

(b) the magnitude of the induced voltage at a constant magnetic field

E = A x B/t

E = (0.302 x 1.61) / 3.9

E = 0.1247 V

E = 124.7 mV

Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV

Answer:

Explanation:

1 )

An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.

Since the object is placed at a distance more than twice the focal length , its image will be inverted , real  and will be of the size less than the size of object . So option B is applied .

B)  real, inverted and smaller than the object.

2 )

An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.

The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C   or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .

B)  real, inverted and smaller than the object.

Answer:

Around the center of the mirror

Answer:

 f = - 20 cm

Explanation:

This exercise asks us for the focal length, which for a lens in air is

                  1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)

where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.

In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression

                 - 1/40 = (n-1) (1 / R₁ -1 / R₂)

                (1 / R₁ - 1 / R₂) = - 1/40 (n-1)

let's calculate

               (1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)

               (1 / R₁ -1 / R₂) = -1/20

 Now we change the construction material for one with refractive index

n = 2, keeping the radii,

              1 / f = (n-1) (1 / R₁-1 / R₂)

              1 / f = (n-1) (-1/20)

let's calculate

             1 / f = (2.00-1) (-1/20)

              1 / f = -1/20

              f = - 20 cm

Answer:

The wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm

d. 198 mm

Explanation:

Refractive index is given by;

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}}[/tex]

where;

[tex]\lambda_{vacuum}[/tex] is the wavelength of the light beam in vacuum

[tex]\lambda_{medium}[/tex] is the wavelength of the beam in a material

[tex]\mu= \frac{\lambda_{vacuum}}{\lambda _{medium}} \\\\\lambda_{vacuum} = \mu *\lambda _{medium}\\\\\ the \ wavelength \ of \ the \ light \ beam \ is \ constant \ in \ a \ vacuum\\\\ \mu_1 *\lambda _{medium}_1 = \mu_2 *\lambda _{medium}_2\\\\\lambda _{medium}_2 = \frac{ \mu_1 *\lambda _{medium}_1 }{ \mu_2} \\\\\lambda _{medium}_2 =\frac{1.5*330}{2.5} \\\\\lambda _{medium}_2 = 198 \ mm[/tex]

Therefore, the wavelength of the ligt beam in a material of refractive index 2.50 is 198 mm.

d. 198 mm

Answer:

The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Explanation:

Given;

energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J

speed of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 x 10⁻³⁴ J.s

The wavelength of the emitted light will be calculated by applying energy of photons;

[tex]E = hf[/tex]

where;

E is the energy emitted light

h is Planck's constant

f is frequency of the emitted photon

But f = c / λ

where;

λ is the wavelength of the emitted photons

[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]

λ ≅ 532 nm

the wavelength of the emitted photons is 532 nm.

Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Answer:

this raise the temperature is x = 3

Explanation:

Heat capacity is the relationship between heat and temperature change

          C = Q / ΔT

if the heat in the system is given by the change in energy and we carry this differential formulas

          [tex]c_{v}[/tex] = dE / dT

In this problem we are told that the energy of thermal radiation is

        E = A V T⁴

Let's look for the specific heat

        c_{v} = AV 4 T³

the power to which this raise the temperature is x = 3

Answer:

Explanation:

The fundamental frequency f₀ is expressed as f₀ =V/2L where;

V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]

m is the mass of the string

L is the length of the string

T is the tension in the string

f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

Given datas

m = 5.69g = 0.00569 kg

T = 440N

L = 0.630 m

Required

Fundamental frequency of the steel piano wire f₀

[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]

Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz