In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gentle 1.2 cm/s current flowing in the direction that the swimmers were going, while in lane 8 there was a current of the same speed but directed opposite to the swimmers' direction. Suppose a swimmer could swim the 50.0 m in 25.0 s in the absence of any current.

Part A: How would the time it took the swimmer to swim 50.0 m change in lane 1?

Part B: How would the time it took the swimmer to swim 50.0 m change in lane 8?

Complete Question

The diagram for this question is shown on the first  uploaded image

Answer:

The initial angular momentum is  [tex]L_i= 0.134 Kg .m^2/s[/tex]

Explanation:

From the question we are told that

        The mass of the rod is [tex]M = 3.5 kg[/tex]

         The mass of the ball is [tex]m =45g = \frac{45}{1000} = 0.045kg[/tex]

          The speed is [tex]v = 5.02 m/s[/tex]

          The angle the ball strikes the rod is [tex]\theta = 42^o[/tex]

           The distance from the center of the rod is [tex]D = \frac{2}{3} L[/tex]

          The length L is  [tex]= 1.2m[/tex]

          The initial angular momentum of the ball is mathematically represented as

                  [tex]L_i = m\ v\ D\ cos \theta[/tex]

Substituting the value

                 [tex]L_i = 45*10^{-3} * 5.02 * \frac{2}{3} * 1.2 * cos (42)[/tex]

                      [tex]= 0.134 Kg .m^2/s[/tex]

Following are the calculation of the initial angular momentum of the ball:

Given:

[tex]M = 3.5 \ kg \\\\m = 45\ g = 0.045\ kg\\\\ v = 5.02\ \frac{m}{s}\\\\ \theta = 42^{\circ}\\\\D = \frac{2}{3}\ L\\\\ L = 1.2\ m\\\\[/tex]

To find:

Li=?

Solution:

Using formula:

[tex]Li= m v D \cos \theta=mv\times \frac{2L}{3}\cos \theta\\\\[/tex]

      [tex]=0.045 \times 5.02 \times \frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.2259 \times\frac{2 \times 1.2}{3}\cos 42^{\circ}\\\\=0.18072\times \cos 42^{\circ}\\\\=0.13430113286\\\\=13.43 \times 10^{-2} \ \frac{kg\ m^2}{s}\\\\[/tex]

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Answer:

20 ohm

Explanation:

V = I x R

R = V/ I

= 120/6

R = 20 ohm

Considering the Ohm's law, the resistance of the coffee machine is 20 Ω.

The current (I) is a measure of the speed at which the charge passes a given reference point in a specified direction. Its unit of measure is amps (A).

The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V). That is, voltage is a measure of the work required to move a charge from one point to another.

Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).

Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.

This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents:

I= V÷R

Where:

In this case, you know that:

Replacing in the Ohm's Law:

6 A= 120 V÷R

Solving:

6 A× R= 120 V

R= 120 V÷6 A

R= 20 Ω

Finally, the resistance of the coffee machine is 20 Ω.

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Answer:

v = 1.95*10^5 m/s

3.13 x 10^16 proton

Explanation:

Identify the unknown:  

The speed of the protons  

List the Knowns:  

Energy of protons KE = 2 keV = 2*10^3 eV

Current produced by the generator: I= 5.05 mA = 5 x 10^-3 A  

1 eV = 1.6 x 10-19 Joule

Mass of proton: m = 1.67 x 10^-27 kg

Charge of proton: q_p = 1.6 x 10^-19 C  

Set Up the Problem:

Kinetic energy is given by:

KE= 1/2mv^2

v = √2KE/m

Solve the Problem:  

v = √2 x 2*10^3 x 1.6 x 10^-19/ 1.67 x 10^-27

v = 1.95*10^5 m/s

b.  Identify the unknown:  

Number of protons produced each second  

Set Up the Problem:  

Current is given by:

I =ΔQ/Δt

So, the total charge in one second:  

ΔQ =I*Δt = 5 x 10^-3 x 1 = 5 x 10^-3 C  

Number of protons in this charge:

n =   ΔQ/q_p

Solve the Problem:  

n =   5 x 10^-3/1.6 x 10^-19  

  =3.13 x 10^16 proton

Answer:

Explanation:

velocity of light in a medium of refractive index V = V₀ / μ

V₀  is velocity of light in air and μ is refractive index of light.

time to travel in tube with air =  length of tube / velocity of light

8.72 ns = L / V₀  L is length of tube .

time to travel in tube with jelly =  length of tube / velocity of light

8.72+ 1.82  = L / V  L is length of tube .

10.54 ns = L / V

dividing the equations

10.54 / 8.72 = V₀  / V

10.54 / 8.72 =   μ

1.21 =  μ

refractive index of jelly = 1.21 .

Answer:

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Explanation:

The magnitude of the magnetic field 'B' can be determined by;

             B = [tex]\frac{mV}{qR}[/tex]

where: m is the mass of proton, V is its speed , q is the charge of proton and R is the distance between the plates.

Given that: speed 'V' of the proton = 3.5 × [tex]10^{6} ms^{-2}[/tex], distance 'R' between the plates = 0.23m, the charge 'q' on proton = 1.9 × [tex]10^{-19}[/tex] C and mass of proton = 1.67 × [tex]10^{-27}[/tex]Kg.

Thus,

  B = (1.67 × [tex]10^{-27}[/tex] × 3.5 ×[tex]10^{6}[/tex]) ÷ (1.6 × [tex]10^{-19}[/tex] ×  0.23)

      = [tex]\frac{5.845 * 10^{-21} }{3.68 * 10^{-20} }[/tex]

      = 0.15883

 B = 0.16 Tesla

The magnitude of the magnetic field 'B' is 0.16 Tesla.

Final answer:

To determine the resultant internal loadings at the cross section at E, we calculate the reaction forces at points B and C. The reaction force at B is 360 lb directed towards the left due to the thrust bearing, while the reaction force at C is 830 lb directed towards the right due to the journal bearing. Combining these forces, the resultant force at E is 1190 lb directed towards the right.

Explanation:

To determine the resultant internal loadings acting on the cross-section at E, we need to consider the forces acting on the shaft at points B and C. Part a requires determining the reaction force at B due to the thrust bearing, part b requires determining the reaction force at C due to the journal bearing, and part c requires combining the forces to find the resultant internal loadings at E.

For part a, since the shaft is supported by a smooth thrust bearing at B, the reaction force at B will be equal in magnitude and opposite in direction to the applied force P1. Therefore, the reaction force at B is 360 lb directed towards the left.

For part b, since the shaft is supported by a journal bearing at C, the reaction force at C will be equal in magnitude to the applied force P2. Therefore, the reaction force at C is 830 lb directed towards the right.

For part c, to find the resultant force at E, we need to combine the forces at B and C. Since the forces are along the same line of action, we can simply add the magnitudes of the forces. The resultant force at E is equal to the sum of the reaction forces at B and C, which is (360 lb) + (830 lb) = 1190 lb directed towards the right.

Answer:

[tex]7.5x10^{14}Hz[/tex], [tex]9.37x10^{14}Hz[/tex].

Explanation:

The ultraviolet light belongs to the electromagnetic spectrum.

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

Any radiation from the electromagnetic spectrum has a speed of [tex]3x10^{8}m/s[/tex] in vacuum.

Therefore, in order to know the frequency the following equation can be used:

[tex]c = \nu \cdot \lambda[/tex]  (1)

[tex]\nu = \frac{c}{\lambda}[/tex]  (2)

Where [tex]\nu[/tex] is frequency, c is the speed of light and [tex]\lambda[/tex] is the wavelength.

Notice that it is necessary to express the wavelength in units of meters before it can be used in equation 2.  

[tex]\lambda = 320nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]3.2x10^{-7}m[/tex]

[tex]\nu = \frac{3x10^{8}m/s}{3.2x10^{-7}m}[/tex]    

[tex]\nu = 9.37x10^{14}s^{-1}[/tex]    

But [tex]1Hz = s^{-1}[/tex]

[tex]\nu = 9.37x10^{14}Hz[/tex]    

[tex]\lambda = 400nm . \frac{1m}{1x10^{9}nm}[/tex] ⇒ [tex]4x10^{-7}m[/tex]

[tex]\nu = \frac{3x10^{8}m/s}{4x10^{-7}m}[/tex]    

[tex]\nu = 7.5x10^{14}Hz[/tex]    

Hence, the frequency range of UVA is [tex]7.5x10^{14}Hz[/tex]  to [tex]9.37x10^{14}Hz[/tex].

Answer:

A) Air resistance acts in a direction opposite the the fall of an object reducing it by doing work against the weight of the object due to gravity.

B) using a ping pong ball, the time of fall will be greatly reduced since it has little weight (its mass x acceleration due to gravity) against the air resistance. The net downward force of the weight and the air resistance will be small.

C) No, I wouldn't expect them to fall at the same time. The steel ball will have more weight compared to the ping pong ball and hence it will have a larger net force downwards.

D) If they are both released from a 6 m height, the steel ball will fall to the ground first since it has a larger net force downwards.

Answer:

The total length is 0.65m.

Explanation:

The total length [tex]x_{tot}[/tex] of the spring is equal to its length [tex]x_0[/tex] right now (0.50 m ) plus the amount [tex]x[/tex] by which it is compressed due to weight of the man:

[tex]x_{tot} = x_0+x[/tex]

The spring compression is given by Hooke's law:

[tex]F = -kx[/tex]

which in our case gives

[tex]-Mg = -kx[/tex]

solving for [tex]x[/tex] we get:

[tex]x= \dfrac{Mg}{k }[/tex]

putting in [tex]M = 150kg, g= 10m/s^2[/tex] and [tex]k = 10,000N/m[/tex] we get:

[tex]x= \dfrac{(150kg)(10m/s^2)}{10,000N/m^2}[/tex]

[tex]x = 0.15m[/tex]

Hence, the total length of the spring is

[tex]x_{tot} = 0.50m+0.15m[/tex]

[tex]\boxed{x_{tot} = 0.65m.}[/tex]

Answer:

Nuclear engineers require a degree in physics in order to have the job.

Explanation:

Answer:

if we double the distance the energy stored will be doubled also

Explanation:

The energy stored in a capacitor is given as

Energy stored =1/2(cv²)

Or

= 1/2(Qv)

Where c = capacitance

Q= charge

But the electric field is expressed as

E= v/d

where v= voltage

d= distance

v=Ed

Substituting into any equation above say

Energy stored =1/2(Qv)

Substituting v=Ev

Energy stored =1/2(QEd)

From the equation above it shows that if we double the distance The energy stored will be doubled also

The issue of farsightedness or presbyopia can be corrected by a converging lens with a focal length of 34 cm and power of 2.94 Diopters.

The provided problem can be solved using the lens formula which is 1/f = 1/v - 1/u, where f = focal length, v = image distance, and u = object distance. The image distance (di) for clear vision should match the lens-to-retina distance, which is constant and taken as 2 cm for the given question. The lens that the physicist is recommended to use should create an image with a reading distance of 25 cm and therefore this will be ‘v’. However, the lens is worn 2 cm in front of the eye and hence the actual image distance becomes 27 cm (or 0.27 m). The object is however at a distance of 97 cm (0.97 m) as indicated by the problem. When we plug these values into the lens formula, we get f = 1 / {(1/0.27) - (1/0.97)} which is approximately equal to 0.34 m or 34 cm. The optical power of lens (P) is reciprocal of the focal length in meters. Hence P = 1/f = 1/0.34 = 2.94 D.

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To correct the physicist's presbyopia for clear vision at 25 cm, a lens with a focal length of approximately -30.35 cm (or -0.3035 m) and a power of around -3.29 diopters is needed. The lens formula helped in determining these values.

Correcting Farsightedness

A presbyopic eye has difficulty seeing objects clearly at close range. In this case, the physicist's eye can only see clearly beyond 97 cm. To correct this and allow clear vision at 25 cm, we need to determine the focal length and power of the corrective lens when worn 2 cm in front of the eye.

Using the lens formula:

1/f = 1/v - 1/u

where:

Substituting the values, we get:

1/f = 1/95 - 1/23

1/f = (23 - 95) / (95 * 23)

1/f = -72 / 2185

f = -2185 / 72 ≈ -30.35 cm

Therefore, the focal length of the lens is approximately -30.35 cm (or -0.3035 m).

The power of the lens (P) is given by the formula:

P = 1/f (in meters)

Substituting the focal length:

P = 1 / -0.3035 ≈ -3.29 diopters (D)

Hence, the power of the corrective lens needed is approximately -3.29 D.

Answer:

It would raise further south

Explanation:

To those of us who live on earth, the most important astronomical object by far is the sun. It provides light and warmth. Its motions through our sky cause day and night, the passage of the seasons, and earth's varied climates.

On any given day, the sun moves through our sky in the same way as a star. It rises somewhere along the eastern horizon and sets somewhere in the west. If you live at a mid-northern latitude (most of North America, Europe, Asia, and northern Africa), you always see the noon sun somewhere in the southern sky.

The sun's path through the rest of the sky is similarly farther north in June and farther south in December. In summary:

In late March and late September (at the "equinoxes"), the sun's path follows the celestial equator. It then rises directly east and sets directly west. The exact dates of the equinoxes vary from year to year but are always near March 20 and September 22.

After the March equinox, the sun's path gradually drifts northward. By the June solstice (usually June 21), the sun rises considerably north of due east and sets considerably north of due west. For mid-northern observers, the noon sun is still toward the south, but much higher in the sky than at the equinoxes.

After the June solstice, the sun's path gradually drifts southward. By the September equinox, its path is again along the celestial equator. The southward drift then continues until the December solstice (usually December 21), when the sun rises considerably south of due east and sets considerably south of due west.

Answer:

It would rise further south

Explanation:

Final answer:

The minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).

Explanation:

In order to suspend the bead in equilibrium, the radiation pressure from the laser beam must be equal to the gravitational force on the bead.

The radiation pressure is given by P = 2I/c, where P is the pressure, I is the laser intensity, and c is the speed of light. The gravitational force is given by F = (4/3)πr³ρg, where F is the force, r is the radius of the bead, ρ is the density, and g is the acceleration due to gravity.

Equating the radiation pressure and gravitational force, we have 2I/c = (4/3)πr³ρg. Rearranging this equation, we get I = (3c/8πr³ρg).

Therefore, the minimum laser power required to suspend the bead in equilibrium is (3c/8πr³ρg).

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m[/tex]

or

[tex]\lambda=8.67\ mm[/tex]

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

Answer:

D from high to low areas

Answer:

The answer is A, friend.

Answer:

11.206 km/s

Explanation: to leave the surface of the earth, you velocity Ve must be;

Escape velocity Ve = (2gRe)^0.5

Where g = acceleration due to gravity 9.81 m/s^2

Re = earth's radius 6400 km = 6.4x10^6

Ve = (2 x 9.81 x 6.4 x 10^6)^0.5

Ve = (125568000)^0.5

Ve = 11205.7 m/s

= 11.206 km/s

The critical area density for a mirror in space where radiation pressure cancels out gravitational attraction can be computed using the principles of radiation pressure and gravitational force. By equating the momentum transfer from reflected sunlight to the gravitational pull of the Sun, one can find the area density at which these forces balance.

The question posed is about finding the critical area density for a mirror located in space at a distance R from the Sun, where the radiation pressure from sunlight would balance the gravitational attraction exerted by the Sun. The solution to this problem involves using the principle of momentum transfer from sunlight and equating it to the gravitational force to derive the density at which the forces are balanced.

The radiation pressure P exerted by sunlight can be calculated using the formula P = 2I/c, where I is the intensity of sunlight and c is the speed of light. Given that sunlight above Earth's atmosphere has an intensity of 1.30 kW/m², the radiation pressure P would be twice this value divided by the speed of light, due to the reflection phenomenon (momentum is doubled as the direction is reversed).

On the other hand, the gravitational force acting on an object is given by the formula F = GMm/R², where G is the gravitational constant, M is the mass of the Sun, m is the mass of the object (or spacecraft with the mirror), and R is the distance from the Sun. The mass m can be represented in terms of the area density ρA, where ρ is the area density and A is the area of the mirror.

Setting the radiation pressure equal to the gravitational force and solving for ρ will yield the critical value of area density at which the two forces cancel out.

The critical value of area density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun is:

[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]

To find the critical value of area density (σ) for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun, we need to set the radiation pressure equal to the gravitational force per unit area.

Let's denote the radiation pressure as [tex]\( P_{rad} \)[/tex] and the gravitational force per unit area as [tex]\( P_{grav} \)[/tex]

The radiation pressure [tex]\( P_{rad} \)[/tex] can be calculated using the formula:

[tex]\[ P_{rad} = \frac{I}{c} \][/tex]

where [tex]\( I \)[/tex] is the intensity of the sunlight at the mirror's position and [tex]\( c \)[/tex] is the speed of light.

The intensity [tex]\( I \)[/tex] at a distance [tex]\( R \)[/tex] from the sun can be found using the inverse square law:

[tex]\[ I = \frac{P_{sun}}{4\pi R^2} \][/tex]

where [tex]\( P_{sun} \)[/tex] is the total power output of the sun.

The gravitational force per unit area [tex]\( P_{grav} \)[/tex] is given by:

[tex]\[ P_{grav} = \frac{F_{grav}}{A} = \frac{G M_{sun} m_{mirror} / R^2}{A} \][/tex]

where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M_{sun} \)[/tex] is the mass of the sun, [tex]\( m_{mirror} \)[/tex] is the mass of the mirror, and [tex]\( A \)[/tex] is the area of the mirror.

Since [tex]\( m_{mirror} = \sigma A \), where \( \sigma \)[/tex] is the area density of the mirror, we can write:

[tex]\[ P_{grav} = \frac{G M_{sun} \sigma A / R^2}{A} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Now, we set [tex]\( P_{rad} = P_{grav} \)[/tex] to find the critical value of [tex]\( \sigma \)[/tex]:

[tex]\[ \frac{I}{c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Substituting [tex]\( I \)[/tex] from the intensity equation, we get:

[tex]\[ \frac{P_{sun}}{4\pi R^2 c} = \frac{G M_{sun} \sigma}{R^2} \][/tex]

Solving for [tex]\( \sigma \)[/tex], we find:

[tex]\[ \sigma = \frac{P_{sun}}{4\pi R^2 c} \cdot \frac{1}{G M_{sun}} \][/tex]

[tex]\[ \sigma = \frac{P_{sun}}{4\pi G M_{sun} R^2 c} \][/tex]

Complete question:- Suppose that the mirror described in Part

Answer:

The magnitude of the acceleration  is [tex]a = 0.33 m/s^2[/tex]

The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis

Explanation:

 From  the question we are that

       The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]

         The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]

         The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]

        The the magnetic field is  [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]

The charge experienced  a force which is mathematically represented as

                    [tex]F = q (\= v * \= B)[/tex]

    Substituting value

         [tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]

            [tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]

            [tex]= (-5.966*10^4 N) \r k[/tex]

Note :

           [tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]

Now force is also mathematically represented as

        [tex]F = ma[/tex]

Making a the subject

      [tex]a = \frac{F}{m}[/tex]

   Substituting values

     [tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]

        [tex]= (-0.33m/s^2)\r k[/tex]

        [tex]= 0.33m/s^2 * (- \r k)[/tex]

Answer:

2.46 W

Explanation:

Thermal conductivity k = 53.4 W/m-K

Radius r = 12 mm = 12x10^-3 m

Lenght = 2.5 m

T1 = 361 °C

T2 = 105 ∘C

Area A = ¶r^2 = 3.142 x (12x10^-3)^2

= 0.00045 m^2

Power P = -AkdT/dx

Where dT = 361 - 105 = 256

dx = lenght of heat travel = 2.5 m

P = -0.00045 X 53.4 X (256/2.5)

= -0.024 X 102.4 = -2.46 W

The negative sign indicates that heat is lost from the metal.

The time required to raise the temperature of the ice from -17.2 °C to 0 °C is approximately 6.22 minutes.

To calculate the time required for the temperature of the ice to rise from -17.2 °C to 0 °C, we can use the equation Q=mcΔT, where Q is the heat supplied, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature.

The heat supplied (Q) during this process is used to raise the temperature of the ice without causing a phase change. Given that Q=820J/min and c=2100J/kg⋅K, we rearrange the equation to solve for time (t): t=Q/mcΔT

Substituting the given values (m=0.570kg, ΔT=0°C−(−17.2°C)=17.2°C), we find t≈820J/min/ (0.570kg)(2100J/kg⋅K)(17.2°C) ≈6.22min.

This calculation assumes that all the heat supplied is used to raise the temperature of the ice without considering the heat required for a phase change. To account for the heat of fusion during the phase change from ice to water at 0 °C, an additional calculation would be needed. However, the question specifies raising the temperature to 0 °C, so the heat of fusion is not considered in this particular scenario.

Understanding the principles of heat transfer, specific heat, and phase changes is essential in solving thermodynamics problems. These principles are widely applicable in various scientific and engineering fields, providing insights into the behavior of materials under different conditions.

Answer:

The total permutation = nP2

The inversions = nC2

Explanation:

Please look at the solution in the attached Word file

Given Information:  

Initial speed = v₁ = 13 m/s

time = t = 4.50 sec

Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m

Altitude = h = 30,000 km = 3×10⁷ m

Required Information:  

a) mass of Mongo = M = ?

b) time in hours = t = ?

Answer:  

a) mass of Mongo = M =  8.778×10²⁵ kg

b) time in hours = t = 11.08 h

Explanation:  

We know from the equations of kinematics,

v₂ = v₁t - ½gt²

0 = 13*4.50 - ½g(4.50)²

58.5 = 10.125g

g = 58.5/10.125

g = 5.78 m/s²

Newton's law of gravitation is given by

M = gC²/4π²G

Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.

M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)

M = 8.778×10²⁵ kg

Therefore, the mass of planet Mongo is 8.778×10²⁵ kg

b) From the Kepler's third law,

T = 2π*(R + h)^3/2/(G*M)^1/2

Where R = C/2π

T = 2π*(C/2π + h)^3/2/(G*M)^1/2

T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2

T = 39917.5 sec

Convert to hours

T = 39917.5/60*60

T = 11.08 hours

Therefore, it will take 11.08 hours for the ship to complete one orbit.

To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The mass of Mongo can then be found using Newton's law of universal gravitation. To determine the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we can use Kepler's third law of planetary motion.

a. What is the mass of Mongo?

To determine the mass of Mongo, we can use the equation of motion for the stone thrown upward and calculate the gravitational acceleration on Mongo. The equation for the motion of the stone is given by:

h = ut + 0.5gt^2

Where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. From the given information, we can find that the stone reaches a maximum height and returns to the ground in 4.50 s. Plugging in the values:

0.0 = (13.0 m/s)(4.50 s) + 0.5g(4.50 s)^2

Simplifying, we find:

g = 13.0 m/s / 4.50 s

Then, we can use Newton's law of universal gravitation to find the mass of Mongo:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant, m1 is the mass of the stone, m2 is the mass of Mongo, and r is the radius of Mongo. Since the stone is on the surface of Mongo, the radius is equal to the circumference divided by 2*pi:

r = 2.00×10^5 km / (2 * pi)

Substituting the known values, we can solve for m2:

m2 = (F * r^2) / (G * m1)

Calculating the gravitational force between the stone and Mongo using the known values and substituting them in the formula, the mass of Mongo is determined.

b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

To find the time it takes for the Aimless Wanderer to complete one orbit around Mongo, we need to use Kepler's third law of planetary motion. This law states that the square of the orbital period (T) is proportional to the cube of the orbital radius (r):

T^2 = (4 * pi^2 * r^3) / (G * m)

Where T is the orbital period, r is the orbital radius, G is the gravitational constant, and m is the mass of Mongo. Plugging in the known values, we can solve for T and convert it to hours.

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The effect of a change in the price of new Sony headphones on the equilibrium price of silicone replacement tips can't be accurately calculated from the provided equations without numerical values. An increase in the price of new headphones would likely lead to a surge in demand for replacement parts, thereby increasing their equilibrium price.

The student is asking about the impact of a change in the price of the new Sony headphones on the equilibrium price of replacement tips. The inverse demand function is given by p = pN - 0.1Q, and the inverse supply function is given by p = 2 + 0.012Q. The equilibrium condition is that the quantity demanded equals the quantity supplied, so we set the demand equation equal to the supply equation and solve for Q and p. Now to find the effect of a change in the price of the new headphones on the equilibrium price of the replacement tips, we need to differentiate the equilibrium price with respect to the price of the new headphones. This essentially involves implicit differentiation of the equilibrium condition.

However, due to the nature of the equations, getting a definitive value for dp/dpN would be difficult without numerical values to plug in. As such, the phrase 'dp/dpN is given by nothing' may have been a typographical error or misunderstanding. An increase in the price of new headphones would likely increase demand for replacement parts, thus raising their equilibrium price. But we cannot compute dp/dpN definitively from the provided equations.

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Answer:

F. See if the slope of Velocity versus Time for both directions are equal but opposite.

Explanation: Velocity is the rate of change of position. The slope of a distance or position-time graph is velocity. Acceleration is the rate of change of velocity. The slope of a velocity-time graph is acceleration.

A velocity-time graph shows changes in the velocity of a moving object over time. The slope of a velocity-time graph represents an acceleration of the moving object.

So, the value of the slope at a particular time represents the acceleration of the object at that instant.

Answer:

Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot

Explanation:

Given

Pressure applied [tex]= 1700[/tex] pounds per square foot

Initial Volume [tex]= 5[/tex] cubic feet

Final Volume [tex]= 9[/tex] cubic feet

As we know pressure is inversely proportional to V

[tex]P = \frac{k}{V}[/tex]

where k is the proportionality constant

V is the volume and

P is the pressure

Work done

[tex]\int\limits^{V_2}_{V_1} {P} \, dV[/tex]

 [tex]\int\limits^{V_2}_{V_1} {\frac{k}{V} } \, dV\\= \int\limits^{V_2}_{V_1} {\frac{1700 * 5}{V} } \, dV\\= 8500* \int\limits^{V_2}_{V_1} {\frac{1}{V} } \, dV[/tex]

Integrating the above equation, we get-

[tex]8500 ln \frac{V_2}{V_1} \\8500 * 2.303 * \frac{9}{5} \\= 4996.18[/tex]

Work done by the gas for the given volume and pressure [tex]= 4996.18[/tex] pounds foot

Answer:

[tex]{\rm K} = 2.4\times 10^{-19}~J[/tex]

Explanation:

The electric field inside a parallel plate capacitor is

[tex]E = \frac{Q}{2\epsilon_0 A}[/tex]

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

[tex]F = qE = \frac{qQ}{2\epsilon_0 A}[/tex]

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

[tex]C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60[/tex]

Plugging this identity into the force equation above gives

[tex]F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q[/tex]

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is [tex]1.6 \times 10^{-19}[/tex]

Therefore, the kinetic energy is [tex]2.4\times 10^{-19}[/tex]

Answer:

please read the answer below

Explanation:

We have that both electric field and magnetic field are given by:

[tex]\vec{E}=E_osin(kx-\omega t)\hat{j}\\\\\vec{B}=B_osin(kx-\omega t)\hat{k}[/tex]

I complete with bold words the answers:

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

a. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0 and B0are the magnitudes of the electric and magnetic fields.

2. amplitudes

b. The variable w is called the angular frequency of the wave.

2. angular frequency

c. The variable k is called the wavenumber of the wave.

1. wavenumber

c.

1. E =E0sin(wt)k^

d.

6. x and t

hope this helps!!

Answer:

The amount of charge on the plates increases during this process

Explanation:

In order to maintain a constant voltage across the plates, pushing the plates closer will increase the charge on the plate