PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large container of water with a higher temperature. Which choice correctly explains what happens to the thermal energy of these systems? A)The smaller container of water has more thermal energy than the larger container, and some of that energy is transferred to the warmer water in a process known as heating. B)The larger container of water has more thermal energy and some of that energy is transferred to the colder water in a process known as heating. C)The larger container of water has more heat and thermal energy than the smaller container. Some heat and thermal energy is transferred to the smaller container of water. D)The smaller container of water has more heat and thermal energy than the larger container. Some heat and thermal energy is transferred to the larger container of water.

Answers

Answer 1

Answer:

that best describes the process is C

Explanation:

This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.

Heat absorbed by the smallest container

             Q_c = m ce ([tex]T_{f}[/tex]-T₀)

Heat released by the largest container is

              Q_a = M ce (T_{i}-T_{f})

how

        Q_c = Q_a

       m (T_{f}-T₀) = M (T_{i} - T_{f})

Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.

Of the final statements, the one that best describes the process is C

since it talks about the thermal energy and the heat that is transferred in the process


Related Questions

The non reflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 636-nm light. If the lens glass has an index of refraction of 1.56, what is the minimum thickness of the coating that will accomplish this task

Answers

Answer:

The  minimum thickness is [tex]t = 1.0192 *10^{-7} \ m[/tex]    

Explanation:

From the question we are told that

   The  refractive index is  [tex]n = 1.29[/tex]

    The  wavelength of the light is  [tex]\lambda = 636 \ nm = 636 *10^{-9} \ m[/tex]

    The  refractive index of the glass lens is  [tex]n_g = 1.56[/tex]

   

Generally the condition for destructive interference of a films is  

               [tex]2t = [ m + \frac{1}{2} ] \frac{\lambda}{n}[/tex]

for  minimum m =  0  

       [tex]2t = [ 0 + \frac{1}{2} ] \frac{\lambda}{n}[/tex]

=>    [tex]2t = [ 0 + \frac{1}{2} ] \frac{636 *10^{-9}}{1.56}[/tex]

=>  [tex]t = 1.0192 *10^{-7} \ m[/tex]    

Name the Sl base units that are ilnportant in chem-istry. Give the Sl units for expressing the following:
(a) length.
(b) volume.
(c) mass,
(d) time,
(e) energy,
(f) temperature

Answers

Answer:

Explanation:

SI unit of

length=meter

volume =dm^3

mass =kilogram

time=second

energy= joule

temperature =kelvin

A beak ball is thrown in an arc and in 2.0s its shadow on the ground travels 180 m in a straight line. What is the average speed of the shadow

Answers

Answer: 90 m/s

Explanation:

The formula for speed is distance/time. The distance is 180 m and the time it took for the ball to travel is 2 s.

180 m/2 s = 90 m/s

A 60.5-kg hiker starts at an elevation of 1280 m and climbs to the top of a peak 2570 m high.
(a) What is the hiker's change in potential energy?
(b) What is the minimum work required of the hiker?
(c) Can the actual work done be greater than this? Explain.

Answers

Answer:

A) Change in potential energy is approximately 766 KJ

B) The minimum work required by the hiker is 765 KJ

C). Yes, the actual work can be greater than this

Explanation:

A) The hiker's potential energy can be calculated at the two different elevations, and then subtracted.

P.E at 1280 m = m X g X h = 60.5kg X 9.81 m/s2 X 1280 = 759, 686 J

P.E at 2570 m = m X g X h = 60.5kg X 9.81 m/s2 X 2570 = 1, 525, 307.85 J

Change in P.E = 1, 525, 307.85 J -759, 686 J = 765, 621.85 J = 766 KJ

B) Work which was done = Force X distance moved.

The force can be got from the effect of gravity on the hiker's mass = 60.5 kg X 9.81 = 593.5 Newton.

The distance moved can be obtained by subtracting the two elevations = 2570 -1280 = 1290 m

Work done = 593 X 1290 = 765, 615 J

The minimum work required by the hiker is 765 KJ

C) Yes, the actual work can be greater than this. This is because we can now put into consideration the fact that the hiker is climbing up the elevation against the force of gravity. This will mean that the hiker is actually doing more work than if he covered the distance on flat terrain,

A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark.
a) If a mass m1 = 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2 = 110 g need to be suspended for the system to be in equilibrium?
b) If a mass m1=80g is suspended at the 25cm mark,and a mass m2 =110g is suspended at the 60 cm mark, from what cm mark would a mass m3 = 45 g need to be suspended for the system to be in equilibrium?

Answers

Answer:

a) 800N × 20 cm = 1100N × x cm

16000= 1100x

x= 14.5

therefore it must be placed on the (50 + 14.5)cm mark

= 64.5 cm mark

b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)

20000 = 11000 + 450x

450x = 9000

x = 20 cm

therefore it must be placed on the (50 + 20)cm mark

= 70 cm mark

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

What is meant by principle of moments?

According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.

a) m₁ = 80 g

m₂ = 110 g

r₁ = 30 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂

Therefore, the distance,

r₂ = m₁r₁/m₂

r₂ = 80 x 20/110

r₂ = 14.54 cm

So, the distance at which mass m₂ should be suspended is,

r' = 50 + 14.54

r' = 64.54 cm

b) m₁ = 80 g

m₂ = 110 g

m₃ = 45 g

r₁ = 25 cm

r₂ = 60 cm

According to the Principle of Moments,

m₁r₁ = m₂r₂ + m₃r₃

80 x 25 = (110 x 10) + (45 x r₃)

45 x r₃ = 2000 - 1100

r₃ = 900/45

r₃ = 20 cm

So, the distance at which mass m₃ should be suspended is,

r' = 50 + 20

r' = 70 cm.

Hence,

a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.

b) The distance at which mass m₃(45 g) should be suspended is 70 cm.

To learn more about Principle of Moments, click:

https://brainly.com/question/28977824

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how many atoms of oxygen are there in one molecule of cardon dioxide , if the chemical formula is CO2

Answers

Answer:

2 oxygen 1 carbon

Explanation:

Answer:

2 oxygen, 1 carbon

Explanation:

20 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What is the ratio of the diameter of a single copper wire to d, if its length is also l and it has the same resistance?

Please show work and type out answer.

Answers

Answer:

D = 4.47d

Explanation:

given that

1/R(eq) = 20/R

R(eq) = R/20

also, we know that the formula for resistance is given by the relation

R = pl/A, where A is πd²/4

If we substitute the value of A, we have

R = pl/(πd²/4)

R = 4pl/πd²

Now, we substitute this in the earlier derived equation

R(eq) = (4pl/πd²) / 20

R(eq) = pl/5πd²

To find the resistance of a single wire made of the same material, the resistance is

R(D) = 4pl / πD²

R(eq) = R(D), and thus

pl/5πd² = 4pl/πD²

1/5d² = 4/D²

D² = 20d²

D = √20d²

D = 4.47 d

The force of gravity will make it easier to stop your car if you are going uphill, and more difficult to stop your car if you are going downhill.

Answers

The force of gravity will cause your vehicle to speed up when going downhill, and slow down when going uphill. The energy of motion increases proportionally with the increase in weight, and the energy increases proportionally with the square of the increase in speed. Increase with an increase of your kinetic energy. Gravity decreases your kinetic energy when driving uphill and increases it when driving downhill. Therefore, the force of gravity will make it easier to stop your car if you're going uphill but more difficult to stop your car if you're going downhill.

Which planet orbits in a different plane than all of the others?

Answers

Although they're all 'close', none of the planets orbits in the same plane as any other planet.  They're all in slightly different planes.

The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit).  But Pluto is officially not a planet, so I don't think it's a good answer.

The next greatest inclination compared to Earth's orbit is Mercury.  That one is about 7 degrees.

The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.

A rectangular object was found to have a mass of 1.278 kg and density of 4.98  g/cm3. Suppose that you knew that the length was 47 mm and the width was 61 mm. Using this information, compute the height of the rectangle in cm.

Answers

Answer:

89.6 cm

Explanation:

From the question,

Volume of the rectangular object = Mass/Density.

V = m/D.................. Equation 1

Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³

Substitute into equation 1

V = 1.278/4980

V = 2.57×10⁻⁴ m³.

But,

V = lwh............... Equation 2

Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.

make h the subject of the equation

h = V/lw........... Equation 3

Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.

Substitute into equation 3

h = 2.57×10⁻⁴/(0.047×0.061)

h = 0.896 m

h = 89.6 cm

(A) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other.
(B) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

Answers

Answer:

A) -40° C and -40° F

B) 574.25° K and 574.25° F

Explanation:

see attachment for calculation and explanation

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concerning this process is correct? (There may be more than one correct choice.)
A. The entropy gained by the iron is equal to the entropy lost by the ocean.
B. The ocean gains less entropy than the iron loses.
C. The change in the entropy of the iron-ocean system is zero.
D. The entropy lost by the iron is equal to the entropy gained by the ocean.
E. The ocean gains more entropy than the iron loses.

Answers

Answer:

E. The ocean gains more entropy than the iron loses.

Explanation:

When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .

Hence option E is correct .

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i ​ 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases

Answers

Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:

[tex]a_{c}=\frac{v^{2}}{r}[/tex]

v is linear speed

r is radius of the curve the object in traveling along

For its first lap:

[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]

After a while:

[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]

[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]

Comparing accelerations:

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]

[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]

[tex]a_{c}_{2}=16a_{c}_{1}[/tex]

With linear speed 4 times faster, centripetal acceleration is 16 times greater.

I don’t understand can someone break it down for me

Answers

Answer:

a = (v² – v₀²)/ 2(s – s₀)

Explanation:

v² = v₀² + 2a (s – s₀)

We can make 'a' the subject of the above expression as follow:

v² = v₀² + 2a (s – s₀)

Subtract v₀² from both side

v² – v₀² = v₀² + 2a (s – s₀) – v₀²

v² – v₀² = v₀² – v₀² + 2a (s – s₀)

v² – v₀² = 2a (s – s₀)

Divide both side by (s – s₀)

(v² – v₀²)/ (s – s₀) = 2a

Divide both side by 2

(v² – v₀²)/ (s – s₀) ÷ 2 = a

(v² – v₀²)/ (s – s₀) × 1/2 = a

(v² – v₀²)/ 2(s – s₀) = a

a = (v² – v₀²)/ 2(s – s₀)

a torque of 6Nm is required to accelerate a wheel from rest to 7.5rev/s In a time of 5.0s.find the moment of inertia of the wheel​

Answers

Your question has been heard loud and clear

Torque = 6Nm

Angular acceleration= 7.5 revolutions / second

Moment of inertia = ?

 

Torque = Moment of inertia * Angular acceleration

Therefore , moment of intertia = Torque / Angular acceleration

Moment of inertia = 6 / 7.5 = 0.8 Kgm^2

Moment of inertia of wheel = 0.8 Kgm^2

Thankyou

(4)
The electric field inside an uncharged metal sphere is initially zero. If the sphere is
then charged positively, the field at the center of the sphere will be :
A) zero
B) finite and directed radially inward
C) nearly infinite
D) finite and directed radially outward

Answers

Answer:

Option (A) : Zero

Explanation:

Electric field Intensity inside a metallic body is always ZERO.

Which of the following statements is true regarding electromagnetic waves traveling through a vacuum?
a. All waves have the same wavelength.
b. All waves have the same frequency.
c. All waves have the same speed.
d. The speed of the waves depends on their wavelength.
e. The speed of the waves depends on their frequency.

Answers

Answer:

C. All waves have the same speed.

Explanation:

Wave equation is given as;

V = fλ

where;

V is the speed of the wave

f is the frequency of the wave

λ is the wavelength

The speed of the wave depends on both wavelength and frequency

The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.

Therefore, the correct option is "C"

C. All waves have the same speed.

Why is area a scalar quantity?

Answers

Answer:

Area is a scalar quantity because there is no need of direction to define and also follow the algebraic summation. When we talk about vector there exists a frame of reference with a certain origin. It actually depends on the fact of the physical area, but if that factor changes to a non-directional object such as a rug spread on the floor, you can consider the area or a region as a scalar. (*Scalars are quantities that are fully described by a magnitude (or numerical value) alone.)

Hope this helps!

A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the ratio of their angular velocities,ω1 / ω2 ?

Answers

Answer:

The  ratio is  [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

Explanation:

From the question we are told that

   The first radius is [tex]R_1[/tex]

    The  second radius is  [tex]R_2[/tex]

   

Generally the angular speed of the turntable is mathematically represented as

       [tex]w_1 = \frac{ v_k }{R_1 }[/tex]

Generally the angular speed of the rubber roller is mathematically represented as

        [tex]w_2 = \frac{ v_k }{R_2 }[/tex]

Where [tex]v_k[/tex] is the velocity of both turntable and  rubber roller

So

    [tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]

    [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]

If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged?

Answers

Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Solve for A if F=MA and F=100 and M=25?

Answers

Answer:

[tex] \boxed{ \boxed{ \bold{A = 4}}}[/tex]

Explanation:

Given,

F = 100 , M = 25

Now, let's find the value of A

[tex] \sf{F = MA}[/tex]

plug the values

⇒[tex] \sf{100 = 25 A}[/tex]

Swap the sides of the equation

⇒[tex] \sf{25A = 100}[/tex]

Divide both sides of the equation by 25

⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]

Calculate

⇒[tex] \sf{A = 4}[/tex]

Hope I helped!

Best regards!!

Given:-

Force,F = 100 N

Mass,m = 25 kg

To find out:-

Calculate the acceleration, a ?

Formula applied:-

F = m × a

Solution:-

We know,

[tex] \sf{F = m × a}[/tex]

Substituting the values of mass and acceleration,we get

[tex] \sf\implies \: 100 = 25 \times a[/tex]

[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]

[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object

Answers

Answer:

a). C = b/2   and C = b/4

b).  [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]

c). m = 63.4 kg (approx.)

Explanation:

Ex. 2.4.4

The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.

The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]

or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex]     or   [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] ,    where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]  

The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] ,  where A and Ф are constants.

A is amplitude of  motion

[tex]$ w_0$[/tex] is the angular frequency of motion

Ф is the phase angle.

Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]

or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]

Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]

a).  [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]

b). [tex]$ w_0^2 = k/m $[/tex]  

or  [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]

Ex. 2.4.5

a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]

                                                                    [tex]$ = -kx-bv $[/tex]

    The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]

    or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] ,  angular frequency of the undamped oscillation.

   γ = b/2m  is called the damping coefficient (γ=C)

   [tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]

   for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz

    k = 4 x (3.14)² x (9.8) x 1.1²  = 4.6 x 10² N/m

  For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz

    k = 4 x 9.8596 x 2 x 9.8 x 0.8²  = 5 x [tex]$ 10^7$[/tex] N/m

   [tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]

or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]

γ = b/2 (for 1 kg)  and   γ = b/4 (for 2 kg)

C = b/2   and C = b/4

b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]

 For two particle problem,

   [tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]

[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]

where, μ is the reduced mass.

This time period is same for both the particles.

c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]

        [tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex]  ( approx.)

What amount of heat is required to increase the temperature of 75.0 grams of gold from 150°C to 250°C? The specific heat of gold is 0.13 J/g°C.A. 750 joulesB. 980 joulesC. 1300 joulesD. 1500 joulesE. 2500 joules

Answers

Answer:

B. 980 joules

Explanation:

Given the following data

initial temperature T1= 150 °C

final temperature T2= 250 °C

specific heat of gold c= 0.13 J/g°C

mass of gold m= 75.0 grams

we can use the expression stated below to solve for the quantity of heat

[tex]Q= mc(T2-T1)---------1[/tex]

Substituting our known data into the expression we can solve for the value of Q

[tex]Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules[/tex]

The quantity of heat need to raise the temperature from 150°C to 250°C  is 975 J

Answer:

B. 980 joules

Explanation:

A 18-kg hammer strikes a nail at a velocity of 7.6 m/s and comes to rest in a time interval of 8.3 ms .

Required:
a. What is the impulse given to the nail?
b. What is the average force acting on the nail?

Answers

Answer:

(a) -136.8 Ns.

(b) -1.135 N

Explanation:

(a)

Impulse: This can be defined as the change in momentum.

From the question,

I = mv-mu.................. Equation 1

Where I = impulse, m = mass of the hammer, v = final velocity, u = initial velocity.

Given: m = 18 kg, u = 7.6 m/s, v = 0 m/s (to rest)

Substitute these values into equation 1

I = 18(0)-18(7.6)

I = -136.8 Ns.

(b)

Average force = It.............. Equation 2

Where t = time.

Given: t = 8.3 ms = 0.0083 s.

Average force = -136.8(0.0083)

Average force = -1.135 N

an electric field of magnitude 200 N/C in the positive x- direction. calculate the acceleration in (m/s^2) of a charged particle of mass 1g and charge 1mC that is released from rest in this field? ​

Answers

1/200 that should get your

answer

If an electron in an atom moves from an energy level of 5 to an energy level of 10:____.
a. a photon of energy 5 is absorbed.
b. a photon of energy 15 is absorbed.
c. a photon of energy 5 is emitted.
d. a photon of energy 15 is emitted.

Answers

Answer

Answer:

a. a photon of energy 5 is absorbed

Explanation:

Because when an electron in a lower energy state absorbs energy, in form of photons it moves to higher energy stage in this case 5 photons because it moved from 5 to 10

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volume doubles.

Required:
What is the gas’s final pressure, in atmospheres, if the gas is diatomic?

Answers

Answer:

The pressure is  [tex]P_2 = 4.25 \ a.t.m[/tex]

Explanation:

From the question we are told that

   The initial pressure is [tex]P_1 = 11.2\ a.t.m[/tex]

   The  temperature is  [tex]T_1 = 299 \ K[/tex]

   

Let the first volume be  [tex]V_1[/tex] Then the final volume will be  [tex]2 V_1[/tex]

 Generally for a diatomic  gas

           [tex]P_1 V_1 ^r = P_2 V_2 ^r[/tex]

Here r is the radius of the molecules which is  mathematically represented as

    [tex]r = \frac{C_p}{C_v}[/tex]

Where [tex]C_p \ and\ C_v[/tex] are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     [tex]C_p=7 \ and\ C_v=5[/tex]

=>   [tex]r = \frac{7}{5}[/tex]

=>  [tex]11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )[/tex]

=>   [tex]P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2[/tex]

=>  [tex]P_2 = 4.25 \ a.t.m[/tex]

The final pressure of the gas will be 4.244 atm.

Given information:

The initial pressure of the gas is [tex]P_1=11.2\rm\;atm[/tex].

The initial temperature of the gas is [tex]T_1=299\rm\; K[/tex].

Let the initial volume of the gas be V. So, the final volume will be double or 2V.

The given diatomic gas is expanded adiabatically. So, the equation of the adiabatic process will be,

[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}[/tex]

where [tex]\gamma[/tex] is the ratio of specific heats of the gas which is equal to 1.4 for a diatomic gas.

So, the final pressure [tex]P_2[/tex] can be calculated as,

[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}\\11.2\times V^{1.4}=P_2\times (2V)^{1.4}\\P_2=4.244\rm\;atm[/tex]

Therefore, the final pressure of the gas will be 4.244 atm.

For more details, refer to the link:

https://brainly.com/question/16014998

A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875 Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?

Answers

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

Answer:

a

  [tex]v _{max } = 6.82 \ m/s[/tex]

b

 [tex]a_{max} = 37489.5 \ m/s^2[/tex]

Explanation:

From the question we are told that

   The amplitude is [tex]A = 1.24 \ mm = 1.24 * 10^{-3} \ m[/tex]

   The frequency is  [tex]f = 875 \ Hz[/tex]

   

Generally the maximum speed is mathematically represented as

           [tex]v _{max } = A * 2 * \pi * f[/tex]

=>        [tex]v _{max } = 1.24*10^{-3} * 2 * 3.142 * 875[/tex]

=>        [tex]v _{max } = 6.82 \ m/s[/tex]

Generally the maximum acceleration is mathematically represented as

       [tex]a_{max} = A * (2 * \pi * f)[/tex]

=>      [tex]a_{max} = 1.24*10^{-3} * (2 * 3.142 * 875 )^2[/tex]

=>       [tex]a_{max} = 37489.5 \ m/s^2[/tex]

Suppose the width of your fist is 4.1 inches and the length of your arm is 35.4 inches. Based on these measurements, what will be the angular width (in degrees) of your fist held at arm’s length?

Answers

Answer:

7 degree

Explanation:

given data

width = 4.1 inches

length = 35.4 inches

solution

we consider as per fig

O is mid point of BC

so  OB  = 2.05 inches

and

AB = [tex]\sqrt{OB^2 + OA^2}[/tex]

AB = [tex]\sqrt{2.05^2 + 35.4^2}[/tex]

AB = 35.078 inches

so

[tex]sin \frac{\alpha }{2} = \frac{OB}{AB}[/tex]

[tex]sin \frac{\alpha }{2} = \frac{2.05}{35.078}[/tex]

[tex]\alpha = 7 degree[/tex]

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find c). You have a spring with spring constant k 5 N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction

Answers

Answer:

  b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

               fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

               w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

             

let's call

               w₀² = k / m

               w² = w₀² - b² / 4m²

               b² = (w₀² -w²) 4 m²

Let's find the angular velocities

             w₀² = 5 / 0.1

             w₀² = 50

             w = 2π f

             w = 2π 1

             w = 6.2832 rad / s

we subtitute

               b² = (50 - 6.2832²) 4 0.1²

               b = √ 0.42086

                b = 0.6487 kg / s

The coefficient friction of the mass during the measurement is 0.648 kg/s.

The given parameters;

mass, m = 0.1 kgspring constant, k = 5 N/mfrequency of the mass, F = 1 Hz

During oscillatory motion, friction is directly proportional to speed.

[tex]F_k = -vb[/tex]

where;

b is the coefficient of friction

The angular velocity is given as;

[tex]\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)[/tex]

From the equation above, we will have the following;

[tex]\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50[/tex]

Also, the instantaneous angular speed is calculated as;

[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s[/tex]

From equation (1), the coefficient of friction is calculated as follows;

[tex]\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s[/tex]

Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.

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