A balloon contains 1.21 x 105 L of ideal gas at 265K. The gas is then cooled to 201 K. What is the volume (L) assuming no gas enters or exits the balloon

Answer:

B. constant pressure, isothermal, adiabatic

Explanation:

A quantity of an ideal gas is compressed to half its initial volume.

The process may be adiabatic, isothermal or occurring at constant pressure.

Adiabatic-constant heat

Constant pressure or isobaric

Isothermal or constant temperature

An external agent is a system that does work on a system or a machine.

This external agent applies force , or changes the state of the body it is acting on.

In order of the work required of an external agent, least to greatest

The following processes will be arranged.

constant pressure, isothermal, adiabatic

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

Answer:

Raw materials are most times gotten from the earth through various forms of extraction procedures.

A) Stainless steel utensils is made up of mainly Iron and other elements such as chromium , carbon etc.

B) Cat litter comprises of ceramic products which is made up of clay.

C) Tums brand antacid tablets comprises of calcium carbonate, magnesium hydroxide, aluminum hydroxide and sodium bicarbonate which could be extracted from the earth.

D)Lithium batteries are made up of elements in the earth such as lithium and carbon.

E)Aluminum beverage cans are made up of aluminum extracted from the ground.

Answer:

65.3 Inches tall

Explanation:

If Sammy is 5 feet and 5.3 inches tall, we simply need to convert the feet to inches, and sum the remaining inches from his height to determine his overall height in inches.

So, 5 feet = (12 inches/1foot) * (5 feet) = 60 inches

And 60 inches + 5.3 inches = 65.3 inches.

Hence, Sammy is 65.3 inches tall.

Cheers.

Answer:

[tex]\theta_1 = 0.400^o[/tex]

[tex]\theta_2 =0.378^o[/tex]

Explanation:

From the question we are told that

    The  number of slits per cm is  k =  [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]

    The order of the maxima is  n =  1

    The wavelength are  [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]

The  spacing between the slit is mathematically represented as

           [tex]d = \frac{ 0.01}{k}[/tex]

=>       [tex]d = \frac{ 0.01}{161}[/tex]

=>         [tex]d = 6.211 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is  

        [tex]n\lambda = d \ sin \theta[/tex]

At  [tex]\lambda_1[/tex]

      [tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]

      [tex]\theta_1 = 0.400^o[/tex]

At  [tex]\lambda_2[/tex]

       [tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]

       [tex]\theta_2 =0.378^o[/tex]

Answer:

The percentage is  [tex]k = 12.5 \%[/tex]

Explanation:

From the question we are told that

    The  axis is is  at  [tex]\theta = 45 ^o[/tex]

Generally the of intensity light emerging from the first polarizer is mathematically represented as

                [tex]I_{1} = \frac{I_o}{ 2}[/tex]

Where  [tex]I_o[/tex] is the intensity of unpolarized light

       Now the light emerging from the second polarizer is mathematically represented as

         [tex]I_2 = I_ 1 * cos ^2(\theta )[/tex]

         [tex]I_2 = \frac{I_o}{2} * cos ^2(45 )[/tex]

        [tex]I_2 = \frac{I_o}{2} * \frac{1}{2} = \frac{I_o}{4}[/tex]

  Now the light emerging from the third  polarizer is mathematically represented as      

       [tex]I_3 = I_ 2 * cos ^2(\theta )[/tex]

       [tex]I_3 = \frac{I_o}{4} * cos ^2(45 )[/tex]

      [tex]I_3 = \frac{I_o}{8}[/tex]

Now the percentage of the intensity of light that emerged with respect to the intensity of  the unpolarized light is

      [tex]k = \frac{\frac{I_o}{8} }{I_o } * 100[/tex]

     [tex]k = 12.5 \%[/tex]

The percentage of light that gets through the three successive Polaroid filters is; 12.5%

We are given;

Angle of transmission axis; θ = 45°

Formula for intensity of light from first polarizer is;

I₁ = ¹/₂I₀

Formula for intensity of light from second polarizer is;

I₂ = I₁cos²θ

Formula for intensity of light from third polarizer is;

I₃ = I₂cos²(90 - θ)

Combining the 3 equations;

Put ¹/₂I₀ for I₁ in second formula to get;

I₂ = ¹/₂I₀cos²θ

Put ¹/₂I₀cos²θ for I₂ in third formula to get;

I₃ = ¹/₂I₀cos²θ*cos²(90 - θ)

Plugging in 45° for θ gives;

I₃ = ¹/₂I₀cos²45*cos²(90 - 45)

⇒ I₃ = ¹/₂I₀cos²45*cos²45

⇒ I₃ = ¹/₂I₀cos⁴45

Now, cos 45 in surd form is 1/√2. Thus;

I₃ = ¹/₂I₀(1/√2)⁴

I₃ = ¹/₂I₀(¹/₄)

I₃ = ¹/₈I₀

I₃/I₀ = ¹/₈

I₃/I₀ = 0.125

In percentage form, we have;

I₃/I₀ = 12.5%

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Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]

Frequency of oscillation = 3.15 Hz

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Answer:

c. above the point of unit elasticity.

Explanation:

The elastic portion of the downward-sloping straight-line demand curve lies above the point of unit elasticity. Supply and demand are fundamental concept in economics. The demand curve shows how much of a good people will want at a different prices. The demands curves illustrates the intuition why people purchase a good for a lower price. For the demand curve, the price is always shown on the vertical axis and the demand curve is shown on the horizontal axis. Thus , the quantity demanded increases as the price gets lower. However, the price elasticity of the demand curve varies along the demand curve. This is because there is a key distinction between the gradient and the elasticity. The gradient which is the slope of the line is always the same in the demand curve but elasticity of the demand changes in the percentage of the quantity demand. Therefore, elasticity will vary along the downward-sloping straight - line demand curve. So,  in a downward-sloping straight-line demand curve, the elastic portion is usually above the  point of unit elasticity

Answer:

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Explanation:

given data

frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz

we consider here intensity I = 0.2 W/m²

solution

we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s

and take energy density is E

so here

E × C = I  

E = [tex]\frac{I}{C}[/tex]   ................1

here C = 3 × [tex]10^{8}[/tex] m/s

so photon density is

photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex]     ...............2

photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

The bullet's vertical velocity at time [tex]t[/tex] is

[tex]v=1400\dfrac{\rm m}{\rm s}-gt[/tex]

where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.

At its highest point, the bullet's vertical velocity is 0, which happens

[tex]0=1400\dfrac{\rm m}{\rm s}-gt\implies t=\dfrac{1400\frac{\rm m}{\rm s}}g\approx\boxed{142.857\,\mathrm s}[/tex]

(or about 140 s, if you're keeping track of significant figures) after being fired.

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

[tex](1093.613 \times x) = 45[/tex]

[tex]x = 45 \div 1093.613[/tex]

x = 0.0411480 km

here x represent maximum length for which he can maintain his maximum speed expressed in kilometers.

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Answer:

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

[tex]a_t = ar[/tex]

where;

a is the angular acceleration and

r is the radius of the circular path

[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]

Determine time of the rotation;

[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]

The magnitude of total linear acceleration is given by;

[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Answer:

Explanation:

Let the race be of a fixed distance x

[tex]Average Speed = \frac{Total Distance}{Total Time}[/tex]

Troy's Average speed = 3 miles/hr = x / 0.2 hr

x = 0.6 miles

Abed's Average speed = 0.6 / 0.125 = 4.8 miles/hr

Answer:

The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Explanation:

We are given that A city of Punjab has 15 percent chance of wet weather on any given day.

So, Probability of wet weather = 0.15

Probability of not being a wet weather = 1-0.15 =0.85

We are supposed to find probability that it will take a week for it three wet weather on 3 separate days

Total number of days in a week = 7

We will use binomial over here

n = 7

p =probability of failure = 0.15

q = probability of success=0.85

r=3

Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]

[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]

Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]

Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]

Standard deviation =0.9447

Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm  = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]

[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m

Answer: i think it is d. none of them.

Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.

Answer:

A The father should sit closer to the pivot.

C The longer wrench makes the job easier because less force is needed when there is more distance from the pivot.

A As far from the head of the hammer as possible because this will maximize torque.

D at the opposite side of the seesaw towards the middle

:) gl

Explanation:

If a father and his son want to play on a seesaw then to balance the torque of the seesaw the father should sit near the pivot as he had more weight as compared to his son, while the son should sit a little farther from the pivot point as compared to his father.

Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.

Mechanical advantage = output force(load) /input force (effort)

As given in the problem statement If a father and his son wish to play on a seesaw,

The father should sit close to the pivot because he weighs more than his son, and the son should sit a little farther away from the pivot point than his father. This will help balance the torque of the seesaw.

Thus, the father should sit near the pivot on the one side and the son should sit a little farther from the pivot of a seesaw on the other side.

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Answer:

30cm

Explanation:

assume that the eyes are substantially above the water so that sin(theta) is approximately theta.

( small angle approximation).

The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d

x/d = sin( angle of incidence)

if the apparent depth of the water is h then

x/h = sin( angle of refraction)

and applying snells law

1 sin ( theta air) = 1.33 sin( theta water)

1 * x/h = 1.33 * x/d

d/h = 1.33

or h/d = 1/1.33

h/39 = 1.33

h = 39 /1.33 so that is the apparent depth of the stream assuming:-

1. Your eyes are almost directly overhead

and

2. your eyes are a significant distance above the surface of the water.

x/d = 1.33 x/h

h/d =39/1.3

= 30cm

Answer:

yes, shaolin monks can do. you can find tutorials on Youtb

Answer:

10.8rev

Explanation:

Using

Wf²-wf = 2 alpha x theta

0²- 56.36x56.36/ 2(-20.13) x theta

Theta = 68.09 rad

But 68.09/2π

>= 10.8 revolutions

Explanation:

Answer:

there are over 100 billion stars in our galaxy.

Answer:

0.61 m

Explanation:

The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.

using the relationship

c = fλ

where

c is the speed of light in vacuum = 3 x 10^8 m/s

f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz

λ is the wavelength = ?

λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m

answer to your question is 0.6m

Answer:

C

They allow light to pass only if their directions of polarizations are exactly 90° apart.

Answer:

Explanation:

Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/

α = 26.56°

Laser angle with Normal = 90 - 26.56 = 63.44 °

Assuming a red laser, refractive index in water is 1.331.

Angle of refraction in water is given by:

Ref Ind = Sin i / Sin r

1.331 = Sin 63.44 / Sin r

Sin r = 0.8945 / 1.331 = 0.6721

Angle r = 42.22°

For the path in water:

Tan 42.22 = x / 3.2

x = 2.9m where x is the lateral displacement of the laser ince it hits the water

So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool

Answer:

-0.24diopters

Explanation:

The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)

So 1/do +1/di = 1/f

1/infinity + 1/-4.2 = 1/f

1/f = 1/-4.2 = -0.24diopters

Answer:

E = 1336.71875 J

Explanation:

We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF

Capacitor 2; C2 = 23.0 µF

Capacitor 3; C3 = 29.3 µF

Supply voltage;V = 125 V

Formula for capacitance in series is;

Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3) .......

Thus, equivalent capacitance is;

C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F

Now, the formula for maximum energy stored is;

E = ½ × C(eq) × V²

E = ½ × 0.1711 × 10^(-6) × 125²

E = 1336.71875 J

Answer:

Explanation:

[tex]Force = 25N\\Area = 0.05m^2\\\\Pressure = \frac{Force}{Area}\\ \\Pressure = \frac{25}{0.05}\\\\ Pressure = 500 Pascals[/tex]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

[tex]F_{r}=kv^2[/tex]

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]

Put the value into the formula

[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]

Put the value of k

[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]

[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]

At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]

So, [tex]1600-v^2=0[/tex]

[tex]v=\sqrt{1600}[/tex]

[tex]v=40\ ft/sec[/tex]

Hence, The velocity is 40 ft/sec.