Suppose that video game discs are a normal good. If the income of video game players​ increases, you predict that in the market for video games

To solve this problem it is necessary to apply the concepts related to energy conservation.

In this case the kinetic energy is given as

[tex]KE = \frac{1}{2} mv^2[/tex]

Where,

m = mass

v= Velocity

In the case of heat lost energy (for all 4 wheels) we have to

[tex]Q = mC_p \Delta T \rightarrow 4Q = 4mC_p \Delta T[/tex]

m = mass

[tex]C_p =[/tex] Specific Heat

[tex]\Delta T[/tex]= Change at temperature

For conservation we have to

[tex]KE = Q[/tex]

[tex]\frac{1}{2} mv^2 = 4mC_p \Delta T[/tex]

[tex]\Delta T = \frac{1}{2}\frac{mv^2}{4mC_p}[/tex]

[tex]\Delta T = \frac{1}{2}\frac{(1500)(30)^2}{4(8)(448)}[/tex]

[tex]\Delta T = 47.084\°C \approx 47\°C[/tex]

Therefore the temperature rises in each of the four brake drums around to 47°C

Answer:

6230.49413 J

0 J

0.63998

Explanation:

F = Force = 40 N

[tex]\theta[/tex] = Angle = 52°

Work done is given by

[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]

The work she does on the flight bag is 6230.49413 J

The work done on the flight bag will be the opposite of the work done by the flight attendant

[tex]W=-6230.49413\ J[/tex]

So net work will be

[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]

The net work done on the flight bag is 0 J

Coefficient of friction is given by

[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]

The coefficient of friction is 0.63998

(a)  The work done by the attendant on the flight bag is 6230.49 J.

(b)  The work on the flight bag is 0 J .

(c)  The coefficient of kinetic friction between the flight bag and the floor is 0.35.

Given data:

The weight of flight bag is, W = 70.0 N.

The distance covered by the bag is, d = 253 m.

The magnitude of force exerted on bag is, F = 40.0 N.

The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].

Work done defined as the product of force and distance covered due to applied force.

(a)

The work done on the flight bag is given as,

[tex]W'=F \times dcos\theta[/tex]

Solving as,

[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]

Thus, the work done by the attendant on the flight bag is 6230.49 J.

(b)

The work done on the flight bag will be the opposite of the work done by the flight attendant. So,

W'' = - W

W'' = - 6230.49 J

Then net work done on the flight bag is,

[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]

Thus, the net work on the flight bag is 0 J .

(c)

The expression for the frictional force is given as,

[tex]f = \mu \times W[/tex]

[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,

[tex]f=Fcos\theta[/tex]

So,

[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]

Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.

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Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = ?

t = time passed by the sample  = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

[tex]k=\frac{2.303}{90.3min}\log\frac{400}{100}[/tex]

[tex]k=1.54\times 10^{-2}\text{ min}^{-1}[/tex]

Now we have to calculate the half-life of substance, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=45min[/tex]

Therefore, the half-life of this substance will be, 45 minutes.

Answer:

W= 61.3 N

Explanation:

The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:

Fg =G*ms*me / (res)²

This force, also obeys the Newton's 2nd Law, so we can write the following equation:

G*ms*me*/ (res)² = ms* a = ms*g

We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:

G*ms*me / (res)² = ms*g = W (1)

where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and

res=  4 *re = 4*6.37*10⁶ m.

Replacing all these known values in (1), we get the value of W:

W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N

The satellite's weight at the altitude of its orbit is;

61.31 N

Formula for gravitational force is;

F = GMm/R²

Where;

G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²

m is mass of earth = 5.97 × 10²⁴ kg

M is mass of satellite = 100 kg

Now, we are told that the altitude is 3R above the Earth's surface.

At the Earth's surface, the distance from the Earth's center is R where R is radius of earth.

Thus, total altitude from the Earth's center to the satellite it (3R + R) = 4R

Thus;

F = GMm/(4R)²

Where R is radius of earth = 6371 × 10⁶ m

Thus;

F = (6.67 × 10⁻¹¹ × 100 × 5.97 × 10²⁴)/(4 × 6371 × 10⁶)

F = 61.31 N

Now, from Newton's second law of motion, we know that the force is equal to the weight.

Thus;

Weight = 61.31 N

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Answer:

[tex]f_{e}[/tex]  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ [tex]m_{e}[/tex]

Where M₀ is the magnification of the objective and  [tex]m_{e}[/tex] is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       [tex]m_{e}[/tex] = 25 /  [tex]f_{e}[/tex]

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/[tex]f_{e}[/tex]  

1) Let's look for the focal length of the eyepiece (faith)

     [tex]f_{e}[/tex]  = - L 25 / f₀ M

     M = 400X = -400

     [tex]f_{e}[/tex]  = - 12 25 /0.40 (-400)

     [tex]f_{e}[/tex]  = 1.875 cm

Let's approximate two significant figures

    [tex]f_{e}[/tex]  = 1.9 cm

Final answer:

The focal length of the eyepiece lens in the microscope is approximately -0.01 mm.

Explanation:

To find the focal length of the eyepiece lens, we can use the formula for the magnification of a compound microscope, which is given by:

M = -fobjective/feyepiece

Where M is the overall magnification, fobjective is the focal length of the objective lens, and feyepiece is the focal length of the eyepiece lens.

Given that the overall magnification is 400x and the focal length of the objective lens is 0.40 cm, we can rearrange the formula to solve for the focal length of the eyepiece lens:

feyepiece = -fobjective/M = -0.40 cm/400 = -0.001 cm = -0.01 mm.

Therefore, the focal length of the eyepiece lens is approximately -0.01 mm.

Answer:

Options A and D are correct

Explanation:

The thermal conductivity of a metal is the property of a metal to allow heat flow through it. conductivity is higher in conductors and low in insulators. Thermal conductivity is high in metals due to the metallic bonds that exist in metals and the presence of free electrons within the metal which allow easy flow of heat from one atom to another.From the problem the rod which contains freer electrons will allow more heat to flow easily hence have a higher thermal conductivity.

Thermal conductivity has the formula below;

[tex]k= \frac{QL}{AΔT}[/tex]

From the above equation we can see that thermal conductivity is inversely proportional to A and directly proportional to L. This mean the rod with less area will have a higher thermal conductivity and the rod with a higher length will have higher k. Hence option C i wrong and option D is correct.

For specific heat, its very much different from thermal conductivity. Specific heat is the ability of a material to hold heat while thermal conductivity is the ability of heat to flow through a material.

Answer:

[tex]\dfrac{1}{28.15432}\ s[/tex]

Explanation:

The factor by which the shutter speed is increased is [tex]\dfrac{7.92}{2.8}[/tex]

Exposure time will be increased by

[tex]f=2^{\dfrac{7.92}{2.8}}=7.1037\ s[/tex]

The proper shutter speed is given by

[tex]\dfrac{1}{T}f=\dfrac{1}{200}\times 2^{\dfrac{7.92}{2.8}}=\dfrac{1}{\dfrac{200}{2^{\dfrac{7.92}{2.8}}}}\\ =\dfrac{1}{28.15432}\ s[/tex]

The proper shutter speed is [tex]\dfrac{1}{28.15432}\ s[/tex]

The formula of an elliptical orbit is given by

[tex]r = \frac{A}{1+Bsin\theta}[/tex]

Assuming the given expression that was wrongly typed and whose true function is

[tex]r = \frac{10.71}{1+0.883sin\theta}[/tex]

We could start by deducing that the greatest distance from the sun would be given at the angle

[tex]\theta = \frac{3\pi}{2}[/tex]

For that value the value of [tex]sin\theta=-1[/tex]

[tex]r = \frac{10.71}{1+0.883(-1)}[/tex]

[tex]r = 91.538 AU[/tex]

That is equal to

[tex]r = 91.54Au* (\frac{93*10^6milles}{1AU})[/tex]

[tex]r = 8513[/tex] million miles

Therefore the correct option is B.

Answer:

a) W =  900   J.  b) Q =  3142.8   J . c) ΔU =  2242.8   J. d) W = 0. e) Q =   2244.78   J.  g) Δ U  =  0.

Explanation:

(a) Work done by the gas during the initial expansion:

The work done W for a thermodynamic constant pressure process is given as;

W  =  p Δ V

where  

p  is the pressure and  Δ V  is the change in volume.

Here, Given;

P 1 = i n i t i a l  p r e s s u r e  =  2.5 × 10^ 5   P a

T 1 = i n i t i a l   t e m p e r a t u r e  =  360   K

n = n u m b er   o f   m o l e s  =  0.300  m o l  

The ideal gas equation is given by  

P V = nRT

where ,

p  =  absolute pressure of the gas  

V =  volume of the gas  

n  =  number of moles of the gas  

R  =  universal gas constant  =  8.314   K J / m o l   K

T  =  absolute temperature of the gas  

Now we will Calculate the initial volume of the gas using the above equation as follows;

PV  =  n R T

2.5 × 10 ^5 × V 1  =  0.3 × 8.314 × 360

V1 = 897.91 / 250000

V 1  =  0.0036   m ^3  = 3.6×10^-3 m^3

We are also given that

V 2  =  2× V 1

V2 =  2 × 0.0036

V2 =  0.0072   m^3  

Thus, work done is calculated as;

W  =  p Δ V  = p×(V2 - V1)

W =  ( 2.5 × 10 ^5 ) ×( 0.0072  −  0.0036 )

W =  900   J.

(b) Heat added to the gas during the initial expansion:

For a diatomic gas,

C p  =  7 /2 ×R

Cp =  7 /2 × 8.314

Cp =  29.1  J / mo l K  

For a constant pressure process,  

T 2 /T 1  =  V 2 /V 1

T 2  =  V 2 /V 1 × T 1

T 2  =  2 × T 1  = 2×360

T 2  =  720  K

Heat added (Q) can be calculated as;  

Q  =  n C p Δ T  = nC×(T2 - T1)

Q =  0.3 × 29.1 × ( 720  −  360 )

Q =  3142.8   J .

(c) Internal-energy change of the gas during the initial expansion:

From first law of thermodynamics ;

Q  =  Δ U + W

where ,

Q is the heat added or extracted,

Δ U  is the change in internal energy,

W is the work done on or by the system.

Put the previously calculated values of Q and W in the above formula to calculate  Δ U  as;

Δ U  =  Q  −  W

ΔU =  3142.8  −  900

ΔU =  2242.8   J.

(d) The work done during the final cooling:

The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.

(e) Heat added during the final cooling:

The final process is a isochoric process and for this, the first law equation becomes ,

Q  =  Δ U  

The molar specific heat at constant volume is given as;

C v  =  5 /2 ×R

Cv =  5 /2 × 8.314

Cv =  20.785  J / m o l   K

The change in internal energy and thus the heat added can be calculated as;  

Q  = Δ U  =  n C v Δ T

Q =  0.3 × 20.785 × ( 720 - 360 )

Q =   2244.78   J.

(f) Internal-energy change during the final cooling:

Internal-energy change during the final cooling  is equal to the heat added during the final cooling Q  =  Δ U  .

(g) The internal-energy change during the isothermal compression:

For isothermal compression,

Δ U  =  n C v Δ T

As their is no change in temperature for isothermal compression,  

Δ T = 0 ,  then,

Δ U  =  0.

During the initial expansion, the gas performed 900 J of work, absorbed 3142.8 J of heat, resulting in a change in internal energy of 2242.8 J. In subsequent processes, work and internal energy changes were determined, culminating in an isothermal compression with no internal energy change.

(a) Work done by the gas during the initial expansion:

The work done (W) for a thermodynamic constant pressure process is given by W = P ΔV, where P is the pressure and ΔV is the change in volume.

Given:

P₁ = initial pressure = 2.5 × 10^5 Pa

T₁ = initial temperature = 360 K

n = number of moles = 0.300 mol

The ideal gas equation is PV = nRT, where P is the absolute pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant (8.314 kJ/mol·K), and T is the absolute temperature of the gas.

Calculate the initial volume of the gas:

P₁V₁ = nRT₁

(2.5 × 10^5) × V₁ = 0.3 × 8.314 × 360

V₁ = (0.3 × 8.314 × 360) / (2.5 × 10^5)

V₁ = 0.0036 m³ = 3.6 × 10⁻³ m³

Given V₂ = 2 × V₁:

V₂ = 2 × 0.0036

V₂ = 0.0072 m³

Now, calculate the work done:

W = P ΔV = (2.5 × 10^5) × (0.0072 - 0.0036)

W = 900 J

(b) Heat added to the gas during the initial expansion:

For a diatomic gas, Cp = (7/2)R, where Cp is the molar heat capacity at constant pressure.

Cp = (7/2) × 8.314

Cp = 29.1 J/mol·K

For a constant pressure process, T₂/T₁ = V₂/V₁:

T₂ = (V₂/V₁) × T₁

T₂ = 2 × T₁ = 2 × 360 = 720 K

Heat added (Q) can be calculated as:

Q = nCpΔT = 0.3 × 29.1 × (720 - 360)

Q = 3142.8 J

(c) Internal-energy change of the gas during the initial expansion:

From the first law of thermodynamics Q = ΔU + W, where Q is the heat added or extracted, ΔU is the change in internal energy, and W is the work done on or by the system.

ΔU = Q - W = 3142.8 - 900

ΔU = 2242.8 J

(d) The work done during the final cooling:

The final cooling is a constant volume (isochoric) process, so there is no change in volume (ΔV = 0), and thus the work done is zero.

(e) Heat added during the final cooling:

For an isochoric process, Q = ΔU. The molar specific heat at constant volume is Cv = (5/2)R.

Cv = (5/2) × 8.314 = 20.785 J/mol·K

The change in internal energy and thus the heat added can be calculated as:

Q = ΔU = nCvΔT = 0.3 × 20.785 × (720 - 360)

Q = 2244.78 J

(f) Internal-energy change during the final cooling:

The internal-energy change during the final cooling is equal to the heat added during the final cooling, so ΔU = Q.

(g) The internal-energy change during the isothermal compression:

For isothermal compression, ΔU = 0 since there is no change in temperature (ΔT = 0).

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Answer:

3900000 m/s

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

v = Speed of the galaxy relative to the earth

Observed frequency is

[tex]f'=(1+0.013)f\\\Rightarrow \dfrac{f'}{f}=1.013[/tex]

Here the Doppler relation must be used.

So, observed frequency is given by

[tex]f'=f\dfrac{v+c}{c}\\\Rightarrow \dfrac{f'}{f}=\dfrac{v+c}{c}\\\Rightarrow v=\dfrac{f'c}{f}-c\\\Rightarrow v=(1.013\times 3\times 10^8)-3\times 10^8\\\Rightarrow v=3900000\ m/s[/tex]

The speed of the galaxy relative to the earth is 3900000 m/s

Answer:

111.657596

Explanation:

The expression of volume is given by

[tex]V=1001.93+111.5282+0.64698m^2[/tex]

Partially differentiating the term we get

[tex]\dfrac{\partial V}{\partial x}=\dfrac{\partial (1001.93+111.5282+0.64698m^2)}{\partial x}\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+2\times 0.64698m\\\Rightarrow \dfrac{\partial V}{\partial x}=111.5282+1.29396m[/tex]

m = 0.100

[tex]\dfrac{\partial V}{\partial x}=111.5282+1.29396\times 0.100\\\Rightarrow \dfrac{\partial V}{\partial x}=111.657596[/tex]

The partial molar volume of glucose is 111.657596

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

[tex]x = v_0 t \frac{1}{2} at^2[/tex]

Where,

x= Displacement

[tex]v_0[/tex] = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

[tex]L = \frac{1}{2} a t_1 ^2[/tex]

For the second cart

[tex]2L \frac{1}{2} at_2^2[/tex]

When the tenth car is aligned the length will be 9 times the initial therefore:

[tex]9L = \frac{1}{2} at_3^2[/tex]

When the tenth car has passed the length will be 10 times the initial therefore:

[tex]10L = \frac{1}{2}at_4^2[/tex]

The difference in time taken from the second car to pass it is 5 seconds, therefore:

[tex]t_2-t_1 = 5s[/tex]

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

[tex]\frac{1}{2} = (\frac{t_1}{t_2})^2[/tex]

[tex]t_1 = \frac{t_2}{\sqrt{2}}[/tex]

From the relationship when the car has passed and the time difference we will have to:

[tex](t_2-\frac{t_2}{\sqrt{2}}) = 5[/tex]

[tex]t_2 (\sqrt{2}-1) = 3\sqrt{2}[/tex]

[tex]t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]

Replacing the value found in the equation given for the second car equation we have to:

[tex]\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]

Finally we will have the time when the cars are aligned is

[tex]18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2[/tex]

[tex]t_3 = 36.213s[/tex]

The time when you have passed it would be:

[tex]20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2[/tex]

[tex]t_4 = 38.172[/tex]

The difference between the two times would be:

[tex]t_4-t_3 = 38.172-36.213 \approx 2s[/tex]

Therefore the correct answer is C.

The passage time for the 10th car in a uniformly accelerating train would be less than 5.0 seconds but not as low as any of the options provided (a to e), as each car passes more quickly than the previous one due to constant acceleration.

The question involves an observer timing how long it takes for individual cars of constant acceleration of a train to pass by. If it took 5.0 seconds for the second car to pass the observer, then the time it takes for the 10th car to pass him is merely the time it took for one car to pass because the train is accelerating at a constant rate, and all cars are of the same length. Considering the train is accelerating uniformly, the passage of each car happens more quickly over time.

Giving this knowledge, for the first car to pass, the train would have started at rest, so it would have taken longer than the second car which had the benefit of the train already moving at a certain velocity. Therefore, we can derive that the time for each subsequent car to pass will be less than 5.0 seconds, which questions the multiple-choice options. Based on this logic, we don't have to calculate the exact passing time for the 10th car, as there are no values given to calculate with, but we can infer that none of the times listed would be correct. Each car would be passing quicker than the last, but it would not be instantaneous; it would be a fraction of the 5 seconds but not as low as any of the times listed from a to e.

Answer:

(a) 66.9%

(b) 147.14 kPa

Explanation:

Given:

Elevation of the water tank z = 15 m

Water volume flow rate V = 70 L/s = 0.07 m³/s

Input electric power consumption by the pump, Welec, in = 15.4  kW

Assuming there are no frictional losses in the pipes and changes in kinetic energy, the efficiency of the pump-motor will be;

                                     η =   ΔEmech ÷ Welec,

Where;

η is the overall efficiency

ΔEmech is the workdone to move the water pumped from the lake to a storage tank 15m above

Welec is the Input electric power consumption by the pump

Solving for ΔEmech ;

                         ΔEmech = mgh

                         mass, m = density × volume  

Density of water = 1000 kg/m³

                          m = 1000 kg/m³ ×  0.07 m³/s

                          m = 70 kg/s

            ∴ ΔEmech = 70 × 9.8 × 15

                               = 10.3 kW

Substituting the values of ΔEmech and Welec to calculate the overall efficiency

             η  =  (10.3 kW ÷ 15.4  kW) × 100 %

                   = 0.6688 × 100 %

                   = 66.88 %

                   = 66.9 %

The overall efficiency of the pump-motor unit is = 66.9 %

(b) The pressure difference between the inlet and the exit of the pump is calculated to be;

                                  Pressure = ΔEmech ÷ V  

                                                  = 10.3 ÷ 0.07

                                                  = 147.14 kPa

This question involves the concept of potential energy, pressure difference, and electrical work.

(a) Efficiency of pump-motor unit is "66.9 %".

(b) The pressure difference between the inlet and the exit of the pump is "147.15 KPa".

The efficiency of the pump-motor unit can be given by the following formula:

[tex]\eta = \frac{W_{P.E}}{W_{elect}}[/tex]

where,

Therefore,

[tex]\eta = \frac{\rho Vgh}{t\ W_{elect}}\\\\\eta=\frac{(1000\ kg/m^3)(0.07\ m^3/s)(9.81\ m/s^2)(15\ m)}{15400\ W}\\\\\eta =0.669 = 66.9\ \%[/tex]

The pressure difference between inlet and outlet of the pump can be found using the following equation:

[tex]\Delta P = \rho gh\\\\\Delta P = (1000\ kg/m^3)(9.81\ m/s^2)(15\ m)[/tex]

[tex]\Delta P = 147150\ Pa = 147.15\ KPa[/tex]

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Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is [tex]920\ x\ 10^{-9} m[/tex] farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      [tex]\vartriangle y =\frac{m \lambda L}{d} [/tex]

                      [tex]m=0,\pm 1,\pm 2,\pm 3,.....[/tex]

        We simply replace the values in that equation :

                      [tex]\vartriangle r= m \lambda =2.\ 460\ nm[/tex]

                      [tex]\vartriangle r= 920\ x\ 10^{-9} m[/tex]

         The dot in the center of fringe E is [tex]920\ x\ 10^{-9}m[/tex] farther from the left slit than from the right slit.

By using the general equation for the double-slit maximums, we will get

The equation for the constructive interference in a double-slit experiment is given by:

[tex]y = \frac{m*\lambda*D}{d} [/tex]

Where:

Now let's answer:

A) If the wavelength is decreased, then the numerator is decreased, meaning the separation between consecutive fringes will also be decreased, so the correct option is a.

B) If d is decreased then the denominator decreases, meaning that the distance between consecutive fringes increases, so the correct option is b.

C) is the distance D is decreased, similar like in case A, the numerator decreases, meaning that the correct option is a again.

D) Sadly, as we do not know:

We can't answer this question.

What we should do here, is to compare the distance between the fringe and each slit, that distance will be the hypotenuse of a right triangle with one cathetus equal to D, and the other cathetus equal to y ± d/2, where each sign corresponds to each slit.

Then the difference in the distance will just be:

[tex]\sqrt{(y - d/2)^2 + D^2} - \sqrt{(y + d)^2 + D^2} [/tex]

If you want to learn more about the double-slit experiment, you can read:

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Answer:

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Explanation:

To know which graph is correct, let's discuss the solution to the problem

Initial mechanical energy

      Em₀ = U₀ = m g H

The mechanical energy at the midpoint

     Em₂ = K + U₂

As there is no friction, mechanical energy is conserved

     Em₀ = Em₂

     U₀ = K + U₂

     K = U₀ - U₂

     K = m g (H - y₂)

Indicates that position 2 corresponds to y₂ = H / 2

     K = m g (H –H / 2)

     K = ½ m g H

     K = ½ Uo

Therefore the graph must be

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Answer: B&C

Explanation:

Release point: Ug bar graph only K none

Halfway point: Ug and K are equal bar graph

second option,

Release point: empty graph

Halfway point: Ug down half K up half

Answer:

Option B is correct.

Explanation:

The resistance of a material or wire is represented by the formula below.

R=(ρL)/A

where

Resistance increases with increase in length and resistivity which means options C and D are wrong.

A=(π[tex]d^{2}[/tex])/4

An increase in diameter will result in a proportional increase in area

From the resistance formula, an increase in area will cause a reduction in the resistance of the material.Thus increased diameter of wire will lower the resistance of wire. Option B is correct

Have a great day

Answer:

2.112 MN

Explanation:

Force = mass × acceleration.  Each second, 264 kg of gas is accelerated from 0 to 8 km/s.

F = ma

F = m Δv / Δt

F = 264 kg × (8000 m/s − 0 m/s) / 1 s

F = 2,112,000 kg m/s²

F = 2.112 MN

Round as needed.

The thrust of the rocket, calculated using the formula - Thrust = mass rate of exhaust x exhaust velocity, is 2.112 Mega Newtons.

The thrust produced by a rocket is typically calculated using the formula: Thrust = mass rate of exhaust x exhaust velocity. In this scenario, the rocket is burning fuel at a rate of 264 kg/s (mass rate of exhaust), and the exhaust gas is ejected at a relative speed of 8 km/s.

However, to ensure the units are consistent, we have to convert the relative speed from km/s to m/s, which results in 8000 m/s (exhaust velocity).

Substituting these values into the formula, we get: Thrust = 264 kg/s x 8000 m/s = 2,112,000 N = 2.112 MN.

Therefore, the thrust of the rocket is 2.112 Mega Newtons.

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Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

        L₀ = I w₀

Final

       [tex]L_{f}[/tex] = [tex]I_{f}[/tex]  [tex]w_{f}[/tex]

      L₀ = [tex]L_{f}[/tex]

      I w₀ = [tex]I_{f}[/tex][tex]w_{f}[/tex]

     [tex]w_{f}[/tex]  = I /[tex]I_{f}[/tex] w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed ​​changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

Final answer:

The correct answer to the question of what occurs to a protostar as it contracts due to the 'conservation of angular momentum' is that the protostar rotates more quickly (d). This is analogous to a figure skater pulling their arms in to spin faster and is supported by observations of star-forming regions like the Orion Nebula.

Explanation:

Stars form from clouds of gas and dust. As these clouds collapse under their own gravity, they form protostars, which spin due to the conservation of angular momentum. Conservation of angular momentum dictates that as the radius of a spinning object decreases, its rotation speed must increase to conserve angular momentum. This concept is similar to a figure skater who spins faster when they pull their arms in. Therefore, when a protostar gravitationally contracts within its parent cloud, it rotates more quickly because as it shrinks, the protostar's rate of spin increases to conserve angular momentum.

This increase in rotation speed results in the formation of a spinning accretion disk around the equator, which is easier to observe in some regions, such as the Orion Nebula or the Taurus star-forming region. Observations from telescopes like the Hubble Space Telescope support this understanding of the role of angular momentum in star formation. In conclusion, the correct answer to the question is (d) the protostar rotates more quickly.

We need to apply the definition of Young's double slit experiment. For fringe width of bright and dark fringe we have that

[tex]\beta = \frac{D\lambda}{d}[/tex]

Or expressed in terms of the wavelength we have that

[tex]\lambda = \frac{\beta d}{D}[/tex]

Where,

[tex]\lambda[/tex]= Wavelength

[tex]\beta[/tex]= Fringe width

d = Slit separation

D = Distance between slit and screen

From the ratios given in the equation, we have that as the wavelength decreases, the pattern determined for the diffraction pattern shrinks, which therefore causes all fringes to get narrower.

Final answer:

When the wavelength of monochromatic light decreases, the single-slit diffraction pattern shrinks with fringes getting narrower (option d). A decrease in slit width results in a wider diffraction pattern. More lines per centimeter on a diffraction grating cause bands to spread farther from the central maximum.

Explanation:

When a monochromatic light passes through a narrow slit, it forms a diffraction pattern due to the phenomenon of diffraction. The pattern consists of a series of bright and dark bands. The bright areas are known as maxima, while the dark areas are known as minima.

If the wavelength of the light decreases, the diffraction pattern shrinks, with all the fringes getting narrower (answer choice d). This is because the angle of diffraction is directly related to the wavelength, and as the wavelength becomes shorter, the angle at which light is bent decreases, causing the fringes to become narrower and the overall pattern to shrink.

Similarly, if the width of the slit producing a single-slit diffraction pattern is reduced, the pattern produced changes as well, with the bands spreading out and becoming wider. This is due to the inverse relationship between slit width and the angle of diffraction. A smaller slit width results in a larger spread of the diffraction pattern.

Lastly, if pure-wavelength light falls on a diffraction grating that has more lines per centimeter, the interference pattern will have bands that spread farther from the central maximum because a higher number of slits per unit area increases the diffractive effects.

The magnitude of the friction force acting on the ketchup bottle is 0.84 N, and it is directed opposite to the initial velocity of the bottle.

The friction force acting on the ketchup bottle can be determined using the equation:

Friction force = Mass x Acceleration

Since the bottle comes to rest, its final velocity is 0 m/s and the acceleration can be calculated using the equation:

Acceleration = (Final Velocity - Initial Velocity) / Time

By substituting the given values, we find that the friction force acting on the bottle is 0.30 kg x (-2.8 m/s) / (1.0 m/s^2) = -0.84 N. The negative sign indicates that the friction force is acting in the opposite direction of the initial velocity, which is in the direction of the attendant's hand.

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Answer:

Explanation:

a.)

The magnitude field

[tex]B=\frac{\mu _0I_{enclosed}}{2\pi r}\\\\=\frac{\mu _0(J_ar^2)}{2\pi r}\\\\=\frac{1}{2}(\mu _0J_ar)\\\\(0.5)(4\pi \times 10^{-7})(20A/m^2)(87\times 10^{-3})\\\\=1.0933\times 10^{-6}T[/tex]

b.)

The displacement current

[tex]i_d=\epsilon_0A\frac{dE}{dt}[/tex]

then [tex]\frac{dE}{dt}=\frac{i_d}{\epsilon_0A}\\\\=\frac{J_d}{\epsilon_0}\\\\=\frac{20}{8.85\times 10^{-12}}\\\\=2.26\times 10^{12}V/ms[/tex]

Answer:

c. The net force is vertically downward.

Explanation:

At the top of  the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.

Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously  on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.

So, the net  force, exactly at the top of the loop, is vertically downward.

Answer:

Yes.

Explanation:

A sitting elephant has zero kinetic energy. A flying bird have some kinetic energy due to its motion. Regardless of their size, a moving object has always more kinetic energy than an object at rest.

Answer:

[tex]2491 kg/m^3[/tex]

Explanation:

Suppose g = 9.8 m/s2. When the statue is suspended from the spring scale, the scale reads 28.4 N. This means the mass of that statue is:

[tex]m = N/g = 28.4 / 9.8 = 2.9 kg[/tex]

When the tub is lowered and submerged in water, the scale reads 17N. So the statue is subjected to a force that make the difference of 28.4 - 17 = 11.4N. This equals to the gravity force of water displaced.

[tex]\rho_wVg = 11.4[/tex]

Let water density [tex]\rho_w = 1000kg/m^3[/tex], we can calculate the volume of the water displaced, which is also the volume of the statue:

[tex]V = \frac{11.4}{g\rho_w} = \frac{11.4}{9.8*1000} = 0.00116 m^3[/tex]

The density of the statue is mass divided by its volume:

[tex]\rho = \frac{m}{V} = \frac{2.9}{0.00116} = 2491 kg/m^3[/tex]

This question involves the concepts of density, weight, volume, and buoyant force.

The density of the ceramic statue is "2495.5 kg/m³".

First, we will find out the mass of the statue:

[tex]W = mg\\m=\frac{w}{g}[/tex]

where,

W = hanging weight of statue = 28.4 N

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]m =\frac{28.4\ N}{9.81\ m/s^2}\\[/tex]

m = 2.9 kg

Now, we will find out the volume of the statue. The difference, in weight of the statue upon submerging, must be equal to the buoyant force applied by the water. This buoyant force is equal to the weight of the volume of water displaced, which is equal to the volume of the statue.

[tex]Difference\ in\ weight\ of\ statue=(Density\ of\ water)(Volume\ of\ Statue)g\\28.4\ N-17\ N=(1000\ kg/m^3)(V)(9.81\ m/s^2)\\\\V=\frac{11.4\ N}{(1000\ kg/m^3)(9.81\ m/s^2)}[/tex]

V = 1.16 x 10⁻³ m³

Now, the density of the ceramic is given as follows:

[tex]\rho = \frac{m}{V} = \frac{2.9\ kg}{1.16\ x\ 10^{-3}\ m^3}\\\\\rho=2495.5\ kg/m^3[/tex]

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The attached picture illustrates the buoyant force.

Answer

Mass of bullet (m) = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

finding the launch speed of bullet

Velocity of wooden block + bullet just after impact

= [\tex]\sqrt{2gh}[/tex]

=[\tex]\sqrt{2\times 9.8 \times 0.6}[/tex]

= 3.43 m/s

v₁ be the launch velocity

Applying law of conservation of momentum

0.03 x v₂ = 0.530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² + 2 g h

v₁² = v₂² - 2 gh

v₁² = 60.6 ² - 2 x (-9.8 )x 0.4

v₁ = 60.65 m /s this is launch speed

Final answer:

The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false.

Explanation:

The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m), and the spin quantum number (s).

Since the quantum numbers determine the energy and other properties of the electrons, if two electrons in the same atom have the same set of quantum numbers, they must have different spins in order to obey the Pauli exclusion principle. Therefore, they would have different energies.

To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to

[tex]T = 0.55s[/tex]

Therefore the frequency will be the inverse of the period and would be given as

[tex]f= \frac{1}{T}[/tex]

[tex]f = \frac{1}{0.55}[/tex]

[tex]f = 1.82s^{-1}[/tex]

The ground state energy of the pendulum is,

[tex]E = \frac{1}{2} hv[/tex]

[tex]E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})[/tex]

[tex]E = 6.03*10^{-34}J[/tex]

The ground state energy in eV,

[tex]E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})[/tex]

[tex]E = 3.8*10^{-15}eV[/tex]

The energy difference between adjacent energy levels,

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

[tex]\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)[/tex]

[tex]\Delta E = 12.1*10^{-34}J[/tex]

Answer:

4.192 N

Explanation:

Step 1: Identify the given parameters

Velocity of airflow = 45m/s

air temperature = 20⁰

plate length and width = 1m and 0.5m respectively.

Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]

[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]

Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²

⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is

[tex]R_{eL} =\frac{VXL}{U}[/tex]

[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]

[tex]R_{eL}[/tex] = 3 X10⁶  (flow is turbulent, Re ≥ 500,000)

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is

[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]

[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]

[tex]C_{f}[/tex] = 0.0038

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is

[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]

[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]

[tex]C_{f}[/tex] = 0.0031

The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]

The total drag force is 4.192 N

Answer:

A) The wave equation is defined as

[tex]y(x,t) = A\cos(kx + \omega t)=0.0275\cos(0.0041x + 6.2t)\\[/tex]

Using the wave equation we can deduce the wave number and the angular velocity. k = 0.0041 and ω = 6.2.

The time it takes for one complete wave pattern to go past a fisherman is period.

[tex]\omega = 2\pi f\\ f = 1/ T[/tex]

T = 1.01 s.

The horizontal distance the wave crest traveled in one period is

[tex]\lambda = 2\pi / k = 2\pi / 0.0041 = 1.53\times 10^3~m[/tex]

[tex]y(x = \lambda,t = T) = 0.0275\cos(0.0041*1.53*\10^3 + 6.2*1.01) = 0.0275~m[/tex]

B) The wave number, k = 0.0041 . The number of waves per second is the frequency, so f = 0.987.

C) A wave crest travels past the fisherman with the following speed

[tex]v = \lambda f = 1.53\times 10^3 * 0.987 = 1.51\times 10^3~m/s[/tex]

The maximum speed of the cork floater can be calculated as follows.

The velocity of the wave crest is the derivative of the position with respect to time.

[tex]v(x,t) = \frac{dy(x,t)}{dt} = -(6.2\times 0.0275)\sin(0.0041x + 6.2t)[/tex]

The maximum velocity can be found by setting the derivative of the velocity to zero.

[tex]\frac{dv_y(x,t)}{dt} = -(6.2)^2(0.0275)\cos(0.0041*1.53\times 10^3 + 6.2t) = 0[/tex]

In order this to be zero, cosine term must be equal to zero.

[tex]0.0041*1.53\times 10^3 + 6.2t = 5\pi /2\\t = 0.255~s[/tex]

The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.

[tex]v(x=\lambda, t = 0.255) = -(6.2\times0.0275)\sin(0.0041\times 1.53\times 10^3 + 6.2\times 0.255) = -0.17~m/s[/tex]

(a) The time taken "1.013 s".

(b) Number of waves "0.987 Hz".

(c) Maximum speed "0.1750 m/s".

A further explanation is below.

Given:

(a)

The time taken will be:

→ [tex]T = \frac{2 \pi}{W}[/tex]

      [tex]= \frac{2 \pi}{6.20}[/tex]

      [tex]= 1.013 \ s[/tex]

The covered horizontal distance will be:

→ [tex]\lambda = \frac{2 \pi}{K}[/tex]

     [tex]= \frac{2 \pi}{0.410}[/tex]

     [tex]= 15.3 \ cm[/tex]

(b)

Wave number,

The number of waves per second will be:

→ [tex]f = \frac{1}{T}[/tex]

     [tex]= \frac{1}{1.013}[/tex]

     [tex]= 0.987 \ Hz[/tex]

(c)

The speed in which the wave crest travel will be:

→ [tex]v = f \lambda[/tex]

      [tex]= 15.3\times 0.987[/tex]

      [tex]= 15.1 \ cm/s \ or \ 0.151 \ m/s[/tex]

and,

The maximum speed of the cork floater will be:

→ [tex]v_1 = AW[/tex]

      [tex]=2.75\times 6.20[/tex]

      [tex]= 0.1750 \ m/s[/tex]

Thus the above answers are correct.

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Answer:a

Explanation:

Given

Specific heat capacity of aluminium is twice that of iron.

[tex]c_{Al}=2c_{iron}[/tex]

Also mass of two blocks is equal

Rate of heat added is also same

[tex]\dot{Q_{Al}}=\dot{Q_{iron}}[/tex]

[tex]\frac{Q_{Al}}{t_{Al}}=\frac{Q_{iron}}{t_{iron}}[/tex]

[tex]\frac{mc_{Al}\Delta T}{t_{Al}}=\frac{mc_{iron}\Delta T}{t_{iron}}[/tex]

[tex]\frac{2c_{iron}}{t_{Al}}=\frac{c_{iron}}{t_{iron}}[/tex]

[tex]t_{Al}=2t_{iron}[/tex]

Thus Time taken by aluminium  block will be more