Whats the temperature -15°F in degrees Celsius?

Answer:

d.  The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applied.

Explanation:

Ideal diode acts like an ideal conductor. In case of forward voltage it acts like an ideal conductor. However when it is reverse biased then it behaves like an ideal insulator. You can understand it bu considering a switch. When the voltage is forward then ideal diode acts like a closed switch. When the voltage is reverse biased then ideal diode behaves like an open switch.

        That is why we can say that the ideal diode acts as a short circuit (higher conduction) for forward currents and as an open circuit ( zero conduction) with reverse voltage applied.

Answer:

1. increase

2. remain the same

3. increases

4. decreases

5. remain the same

6. increases

7. decreases

8. remain the same

Explanation:

a. the formula for the capacitance of a capacitor relating the capacitor and the dielectric material is express as

[tex]C=e_oA/d[/tex]........equation 1

also the capacitance and the charge is related as

Q=CV.......equation 2

from equation 1 as the dielectric material is introduced, the capacitance increases, the charge also increases

2. from the equation as the dielectric material is introduced, the capacitance increases, the electric potential also remain the same

3. from

[tex]C=e_oA/d[/tex]........equation 1

we conclude that the capacitance increases

4. the voltage between the plates decreases

5. the charge remain the same because capacitance is constant

6. the capacitance increases

7. the electric potential decreases

8. remain the same

Answer 1: the charge on the plates will increase

Explanation: placing a dielectric between two charged plate increases its capacitance

C = Q/V,

If the plates remain connected then the voltage remains the same.

Therefore for an increase in capacitance charge will increase.

Answer 2: electric field potential remains the same.

Explanation: if the plates are disconnected, charge on plates remains contant while voltage varies with change in distance, electric field intensity remains constant and this is proportional to the electric potential energy.

Answer 3: capacitance increases

Explanation: introducing a dielectric between two plates causes opposite charges to be induced on the faces of the dielectric. This also reduces the p.d across the capacitor.

Answer 4: voltage remains constant.

Explanation: A connected plate has a constant voltage across its field.

Answer 5: charge remains contant.

Explanation: capacitance will increase with introduction of dielectric, p.d across the plates will drop, the charge will remain constant.

Answer 6: capacitance increases

Explanation: placing a dielectric between plates always increase the capacitance.

Answer 7: electric potential energy falls.

Answer 8: voltage between plates decreases

Answer:

a) centripetal acceleration= v^2/r

=475*475/3080=73.154 m/s^2

b) yes the pilot is okay because 73.154 m/s^2 is less than 9g.

Explanation:

Answer:

the acceleration experienced by a pilot flying in a circle is 126m/s²

Explanation:

Given that,

Speed of pilot, v = 475 m/s

Radius of the circle, r = 1790 m

If the pilot is moving in a circular path, it will experience a centripetal acceleration. It is given by the formula as :

The angular acceleration

a= ω²r

= v²/r.

Where v is tangential velocity and r is the radius.

a = v²/r.

[tex]a=\dfrac{v^2}{r}a=\dfrac{(475\ m/s)^2}{1790\ m}a=126\ m/s^2[/tex]

So, the acceleration experienced by a pilot flying in a circle is 126m/s²

Answer:

Which stars are coming toward us? Which are moving away?

The Star A and B are moving toward the observer and Star C and D away from the observer.

Which star is moving fastest relative to us?

Star A is moving fastest relative to us.

What are the speeds of Star B and Star C?

The speed of the star B is 986842m/s and the speed of star C is  740131m/s.

Explanation:

Which stars are coming toward us? Which are moving away?  

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving  toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ([tex]\lambda_{0} = 121.6nm[/tex])

[tex]Redshift: \lambda_{measured}  >  \lambda_{0} [/tex]

[tex]Blueshift: \lambda_{measured}  <  \lambda_{0} [/tex]

Then, for this particular case it is gotten:

Star A: [tex]\lambda_{measured} = 120.5nm[/tex]

Star B: [tex]\lambda_{measured} = 121.2nm[/tex]

Star C: [tex]\lambda_{measured} = 121.9nm[/tex]

Star D: [tex]\lambda_{measured} = 122.9nm[/tex]

Star A:

[tex]Blueshift: 120.5nm  <  121.6nm [/tex]

Star B:

[tex]Blueshift: 121.2nm  <  121.6nm [/tex]

Star C:

[tex]Redshift: 121.9nm  >  121.6nm [/tex]

Star D:

[tex]Redshift: 122.9nm  >  121.6nm [/tex]

Therefore, according to the approach above. The Star A and B are moving toward the observer and Star C and D away from the observer.

Due to that shift the velocity of the star can be determined by means of Doppler velocity.

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex] (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

[tex]v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})[/tex]

Which star is moving fastest relative to us?

Case for the Star A:

[tex]v = (3x10^{8}m/s)(\frac{121.6nm-120.5nm}{121.6nm})[/tex]

[tex]v = 2713815m/s[/tex]

Case for the Star B:

[tex]v = (3x10^{8}m/s)(\frac{121.6nm - 121.2nm}{121.6nm})[/tex]

[tex]v = 986842m/s[/tex]

Hence, Star A is moving fastest relative to us.

What are the speeds of Star B and Star C?

Case for the Star C:

[tex]v = (3x10^8m/s)(\frac{(121.9nm - 121.6nm}{121.6nm})[/tex]

[tex]v = 740131m/s[/tex]

Therefore, The speed of the star B is 986842m/s and the speed of star C is  740131m/s.

Final answer:

Stars with shift towards shorter wavelengths (Star A and B) are moving toward Earth, while those with shift towards longer wavelengths (Star C and D) are moving away, with Star D moving the fastest. The speeds of Star B and Star C can be calculated using the Doppler effect formula. Star A is blue-shifted the most, indicating it is moving towards us the fastest.

Explanation:

The Doppler Effect on Spectral Lines

The observed changes in the wavelength of spectral lines from hydrogen transitions can be explained by Doppler shift, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. Stars showing a shorter wavelength than the rest wavelength are moving towards us (blue-shifted), while those with a longer wavelength are moving away from us (red-shifted).

For star A with an observed wavelength of 120.5 nm, the shift is towards the blue, indicating it is moving towards us. Star B, with an observed wavelength of 121.2 nm, is also moving towards us but at a slower speed. Star C and Star D, having observed wavelengths of 121.9 nm and 122.9 nm respectively, display a red shift, meaning they are moving away from us. Among these stars, Star D is the one moving away the fastest.

To calculate the velocity of Star B and Star C with respect to Earth, we can use the Doppler effect formula for light:

[tex]v = c × (λ_{observed} - λ_{rest}) / λ_{rest}[/tex]

Where v is the velocity of the star, c is the speed of light, [tex]λ_{observed}[/tex] is the observed wavelength, and [tex]λ_{rest}[/tex] is the rest wavelength of the emission line. By plugging in the values and solving for v, we can determine the velocity for each star.

Answer:

Answer explained below

Explanation:

magnetic force of charged particle placed in uniform magnetic field is given by

F = iLB sin theta

where i is current

L is length

B = magnetic field

theta is the angle between L and B

so we can design an experiment with the given apparatus

as we have device for measuring length ( meter stick and scale)

magnitude of magnetic field from bar magnetic using magnetic moments principle

Finally from conducting wires and power, we can measure current i using Ammeter

Back emf is 85.9 V.

Explanation:

Given-

Resistance, R = 3.75Ω

Current, I = 9.1 A

Supply Voltage, V = 120 V

Back emf = ?

Assumption - There is no effects of inductance.

A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.

So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,

The voltage across the resistance would be

v = I x R

  = 9.1 x 3.75

  = 34.125 volts

We know,

supply voltage = back emf + voltage across the resistance

By plugging in the values,

120 V = back emf + 34.125 V

Back emf = 120 - 34.125

                = 85.9 Volts

Therefore, back emf is 85.9 V.

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

[tex]V_{orbital} = \sqrt{\frac{GM_E}{R}}[/tex]

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

[tex]V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}[/tex]

[tex]V_{orbital} = 5591.62m/s[/tex]

[tex]V_{orbital} = 5.591*10^3m/s[/tex]

PART B) The period of satellite is given as,

[tex]T = 2\pi \sqrt{\frac{r^3}{Gm_E}}[/tex]

[tex]T = \frac{2\pi r}{V_{orbital}}[/tex]

[tex]T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}[/tex]

[tex]T = 238.61min[/tex]

PART C) The gravitational force on the satellite is given by,

[tex]F = ma[/tex]

[tex]F = \frac{1}{4} mg[/tex]

[tex]F = \frac{270*9.8}{4}[/tex]

[tex]F = 661.5N[/tex]

(a) The orbital speed of the satellite is 5591.62 m/s.

(b) The period of revolution is 238.47 minutes.

(c) The gravitational force on satellite 661.5 N.

(a) The orbital velocity is given by;

[tex]v_{o}=\sqrt{\frac{GM}{r} }[/tex]

Here, 'M' is the mass of the earth and 'r' is the distance is from the centre of the earth to the satellite.

[tex]v_o = \sqrt{\frac{(6.67 \times 10^{-11})\times (5.972\times 10^{24})}{2\times 6370\times 10^{3}} } =5591.62\,m/s[/tex]

(b) The period of revolution is given by;

[tex]T=\frac{2\pi r}{V_o}=\frac{2\times 3.14\times 2\times 6370\times 10^3}{5591.62} =14308.411\,s = 238.47\,min[/tex]

(c) The gravitational force on the satellite is given by;

[tex]F_g = mg_s[/tex]

Where [tex]g_s[/tex] is the acceleration due to gravity at the satellite's height.

Given that, [tex]g_s = \frac{g}{4}[/tex]

[tex]F_g = \frac{270\times 9.8}{4} =661.5\,N[/tex]

Learn more about orbital motion here:

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Answer:

This is the property of metals like W (Tungsten) to produce a continuous spectrum.

Explanation:

The molecules of a gas are highly differentiated and single atoms absorb photons instead of bulk like metals and only give line spectrum for example line spectrum given by hydrogen etc. On the other hand, metals like Tungsten gets very hot and emission of every possible wavelength of light is observed. This emission is continuous because the wavelengths of light in the spectrum has no breakages.

Answer: The car makes a sudden stop.

Explanation:

Manuel is coasting on his bike. Because he is not pedaling, his bike will come to a stop. Which of these will cause Manuel's bike to stop? - the force of friction.

Hope this helps! enjoy your day.

Answer:

Vx =  10.9 m/s ,  Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s  × cos 55°

Vx =  10.9 m/s

Vx = V sin θ

Vy = 19 m/s  × sin 55°

Vy = 15.6 m/s

15.56m/s and 10.90m/s respectively

The vertical and horizontal components of a given vector, say A, are given by

[tex]A_{Y}[/tex] = A sin θ                  ----------------(i)

[tex]A_{X}[/tex] = A cos θ                 ----------------(ii)

Where;

[tex]A_{Y}[/tex] is the vertical component of the vector A

[tex]A_{X}[/tex] is the horizontal component of the vector A

A is the magnitude of the vector A

θ is the angle the vector makes with the positive x-axis (horizontal direction).

Now, from the question;

The vector is the velocity of the 1-kg discus. Lets call it vector V

The magnitude of the velocity vector V = V = 19m/s

The angle that the vector makes with the positive x-axis = θ

To calculate θ;

Notice that the velocity vector makes an angle of 35° from the vertical direction rather than the horizontal direction.

Therefore, to get the horizontal direction of the velocity vector, we subtract 35° from 90° as follows;

θ = 90° - 35° = 55°

Now, the vertical and horizontal components of the velocity vector, V, are given by

[tex]V_{Y}[/tex] = V sin θ              --------------------(iii)

[tex]V_{X}[/tex] = V cos θ             ------------------------(iv)

Substitute all the necessary values into equations(iii) and (iv) as follows;

[tex]V_{Y}[/tex] = 19 sin 55° = 19m/s x 0.8192 = 15.56m/s

[tex]V_{X}[/tex] = 19 cos 55° = 19m/s x 0.5736 = 10.90m/s

Therefore, the vertical and horizontal velocity components are respectively 15.56m/s and 10.90m/s.

Answer:

d. The spring force and gravity.

Explanation:

The forces that can be associated with a potential energy function are only the conservative forces. These are the forces whose work done on an object by the force does not depend on the path taken, but only on the initial and final position of the object.

The only two conservative forces in this problem are:

- Gravity

- The spring force

While the air resistance is a non-conservative force.

The potential energy associated with the gravitational force is:

[tex]U=mgh[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

h is the position of the object with respect to a reference level (e.g. the ground)

The potential energy associated to the spring force is

[tex]U=\frac{1}{2}kx^2[/tex]

where:

k is the spring constant

x is the elongation of the spring with respect to the equilibrium position

Answer:

(A) three times as large as the initial value.

Explanation:

Boyle law states that the pressure is inversely proportional to volume if we kept the temperature constant.

[tex]P_{1} V_{1} = P_{2} V_{2}[/tex]        (1)

here [tex]P_{1}[/tex] and [tex]V_{1}[/tex] are the initial pressure and volume.

[tex]P_{2}[/tex] and [tex]V_{2}[/tex] are the final pressure and volume respectively.

now

[tex]P_{1} = p\\V_{1} = v\\P_{2} = ?\\V_{2} = \frac{v}{3}[/tex]

by putting these values in equation 1.

hence it is proved that " if the volume is decreased to one third of its original value then the pressure will be three times larger then its initial value".

Answer:

0.36s, 2.3s

Explanation:

Let gravitational acceleration g = 9.81 m/s2. And let the throwing point as the ground 0 for the upward motion. The equation of motion for the rock leaving your hand can be written as the following:

[tex]s = v_0t + gt^2/2[/tex]

where s = 4 m is the position at 4m above your hand. [tex]v_0 = 13 m/s[/tex] is the initial speed of the rock when it leaves your hand. g = -9.81m/s2 is the deceleration because it's in the downward direction. And t it the time(s) it take to get to 4m, which we are looking for

[tex]4 = 13t - 9.81t^2/2[/tex]

[tex]4.905 t^2 - 13t + 4 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{13\pm \sqrt{(-13)^2 - 4*(4.905)*(4)}}{2*(4.905)}[/tex]

[tex]t= \frac{13\pm9.51}{9.81}[/tex]

t = 2.3 or t = 0.36

Explanation:

Below is an attachment containing the solution.

Answer:

The no. of electron in the beam = [tex]1.64\times10^{12}[/tex]

Explanation:

Given :

The diameter of circular ring = 68 m.

The current flowing in the beam = 0.37 A

Speed of light = [tex]3\times10^{8} ms^{-1}[/tex]

We know that the current is equal to the charge per unit time.

⇒    [tex]I = \frac{Q}{t}[/tex]

∴    [tex]Q=It[/tex]

Here given in the question, electron revolving in a circle with the diameter

[tex]d = 68[/tex]m

⇒ Time take to complete one round [tex](t) =[/tex] [tex]\frac{\pi d }{v}[/tex]

∴    [tex]Q = \frac{I\pi d }{v}[/tex]

     [tex]Q = \frac{0.37 \times 3.14 \times 68}{3 \times 10^{8} }[/tex]

     [tex]Q = 26.33 \times 10^{-8}[/tex]

Now, for finding the no. of electron we have to divide [tex]Q[/tex] to the charge of the electron  [tex]q = 1.6 \times 10^{-19}[/tex]

∴     [tex]n[/tex] =  [tex]\frac{26.33 \times 10^{-8} }{1.6 \times 10^{-19} }[/tex]

      [tex]n = 1.64 \times 10^{12}[/tex]

Thus, the no. of electron in the beam is [tex]1.64 \times 10^{12}[/tex].

Final answer:

To calculate the number of electrons in the beam, first determine the total charge passing a point in one second using the current, then divide by the charge of one electron. Using the given current of 0.37 A, this method reveals the total number of electrons in the beam.

Explanation:

To determine the number of electrons in a beam with a current of 0.37 A, the given data of the SPEAR storage ring can be used. First, recall that the charge of one electron is approximately ‑1.602 × 10⁻¹⁹ C (coulombs).

Current (I) is defined as the rate of charge (Q) flow through a given point, over time (t), expressed as I = Q/t. Therefore, the total charge in the beam can be calculated for one second as the product of the current and time.

To find the number of electrons (N), the total charge is divided by the charge of one electron, mathematically represented as N = Q / e. Substituting the given value of 0.37 A for I and using 1 second for t, the calculation would be N = (0.37 C/s) / (1.602 × 10⁻¹⁹ C/electron). This gives the total number of electrons circulating in the beam.

Answer:

[tex]\sigma_i=1.06*10^{-6}C[/tex]

Explanation:

When the space is filled with dielectric, an induced opposite sign charge appears on each surface of the dielectric. This induced charge has a charge density related to the charge density on the electrodes as follows:

[tex]\sigma_i=\sigma(1-\frac{E}{E_0})[/tex]

Where E is the eletric field with dielectric and [tex]E_0[/tex] is the electric filed without it. Recall that [tex]\sigma[/tex] is given by:

[tex]\sigma=\epsilon_0E_0[/tex]

Replacing this and solving:

[tex]\sigma_i=\epsilon_0E_0(1-\frac{E}{E_0})\\\sigma_i=8.85*10^{-12}\frac{C^2}{N\cdot m^2}*3.40*10^5\frac{V}{m}*(1-\frac{2.20*10^5\frac{V}{m}}{3.40*10^5\frac{V}{m}})\\\sigma_i=1.06*10^{-6}C[/tex]

Answer:

Explanation:

The explanation or solution is given in the attach document

Final answer:

To find the voltage and current as functions of time in a circuit, replace current I(t) with C*dV(t)/dt, and substitute this into Ohm's law V(t)=I(t)R to obtain a differential equation V(t) = (C*dV(t)/dt)*R.

Explanation:

To rewrite the relation V(t)=I(t)R by replacing I(t) with an expression involving the time derivative of the voltage, first understand that the current I(t) is the rate of change of charge Q with respect to time, that is I(t) = dQ/dt. However, since Q is related to the voltage across a capacitor by the equation Q=CV(t), where C is the capacitance, the current can also be expressed in terms of the voltage as I(t) = C*(dV(t)/dt). Replacing this into the original equation gives us V(t) = (C * dV(t)/dt) * R, which is a first-order differential equation relating the voltage to its time derivative.

Answer: The charge on the particle is positive

While the magnitude = 0.00028C

Explanation:

Please find the attached file for the solution

Answer:

It's converted into heat.

Explanation:

Most of the rest of the energy is converted into heat, as can be verified by touching an incandescent lamp that has been turned on for a while. This happens because the light itself is generated by heating the filament with an electric current (Joule heating) until it glows.

Answer:

K= 1226.25 N/m

Explanation:

Given:  mass m = 10 kg, Distance x= 8 cm = 0.08 m, g= 9.81 m/s²

By Hook's Law

F=K x

F=W=mg = 10 kg x 9.8 m/s² = 98.1 N

to Find Spring constant k = F/x  = 98 N /0.08 m

K= 1226.25 N/m

Answer:

     v’= 9.74 m / s

Explanation:

The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.

Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer

        f₁ ’= f₀ (v + v₀)/v

Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest

         f₂’= f₁’ v/(v - vs)

Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’

            v’= vo = vs

Let's replace

           f₂’= f₀   (v + v’)/v   v/(v -v ’)

           f₂’= f₀   (v + v’) / (v -v ’)

           (v –v’ ) f₂’ / f₀ = v + v ’

           v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)

           v’ (1 + 1.059) = 340 (1.059 - 1)

           v’= 20.06 / 2.059

           v’= 9.74 m / s

Answer:

X = 15.88 m

Explanation:

Given:

Initial Velocity V₀ = 13.4 m/s

θ = 30.1 °

g = 9.8 m/s²

To Find horizontal distance let "X" we have to time t first.

so from motion 2nd equation at Height h = 0

h = V₀y t + 1/2 (-g) t ²                                (ay = -g)

0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t²           (V₀y = V₀ Sin θ)

⇒  t = 1.37 s

Now For Horizontal distance  X, ax =0m/s²

X = V₀x t + 1/2 (ax) t ²

X = 13.4 m × cos 30.1° x 1.37 s + 0

X = 15.88 m

To find the distance between the points where the stones strike the ground, analyze the motion of each stone separately and find the horizontal displacements.

To find the distance between the points where the stones strike the ground, we can analyze the motion of each stone separately. For the stone thrown below the horizontal, we can use the equations of motion in the x and y directions to find the time of flight and the horizontal displacement. For the stone thrown above the horizontal, we can use the same approach. Finally, we can subtract the two horizontal displacements to find the distance between the points where the stones strike the ground.

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Coefficient of volume expansion is 8.1 ×10⁻⁴ C⁻¹.

Explanation:

The volume expansion of a liquid is given by ΔV,

ΔV = α V₀ ΔT

ΔT = change in temperature  = 48.5° C

α =  coefficient of volume expansion =?

V₀ = initial volume = 2.35 m³

We need to find α , by plugin the given values in the equation and by rearranging the equation as,

[tex]\alpha=\frac{\Delta \mathrm{V}}{\mathrm{V}_{0} \Delta \mathrm{T}}=\frac{0.0920}{2.35 \times 48.5}=0.00081[/tex]

  = 8.1 ×10⁻⁴ C⁻¹.

Answer:

8.1

Explanation:

Answer:

Explanation:

a Downward acceleration of car A along the slope

= g sinθ - μ g cosθ

= g ( sinθ - μ  cosθ)

= 9.8 ( sin 12 - .6 x cos 12 )

= 9.8 x ( .2079 - .5869 )

= - 3.714 m / s²

So there will be deceleration

v² = u² - 2 a s

= 18² - 2 x 3.714 x 24

= 324 - 178

= 146

v = 12 .08 m /s

b )

In the second case , kinetic friction changes

downward acceleration

= g ( sinθ - μ  cosθ)

= 9.8 ( sin12 - .1 x cos 12 )

9.8 ( .2079  -  .0978 )

= 1.079 m /s

there will be reduced acceleration

v² = u² - 2 a s

= 18² +2 x1.079 x 24

= 324 + 52

= 376

v = 19.4  m /s

Final answer:

The speed of car A hitting car B can be determined by applying the principles of conservation of mechanical energy and work-energy theorem with different coefficients of kinetic friction.

Explanation:

The speed of car A hitting car B can be calculated using the principle of conservation of mechanical energy along with the work-energy theorem.

For part (a) with a coefficient of kinetic friction of 0.60, the speed of car A hitting car B is approximately 6.57 m/s.

For part (b) with a coefficient of kinetic friction of 0.10, the speed of car A hitting car B is approximately 6.93 m/s.

Answer:

Explanation:

Answer:

Explanation:

The intensity and the temperature of metal is constant so the number of photo electrons remains constant. As the number of photo electrons remains same so the photo electric remains constant.

As the frequency is increased, the kinetic energy of the photo electrons increases and thus, the speed of photo electrons increases.

When the frequency of the incident light is increased above a certain threshold, electrons will start being ejected from the metal's surface, with their maximum velocity increasing linearly with the frequency.

As the frequency of the incident light is increased beyond a certain threshold frequency, electrons will begin to be ejected from the surface of the metal. However, their kinetic energy will not increase proportionally with the frequency, rather the maximum velocity of the ejected electrons will increase linearly with the frequency of the incident light.

Answer:(2) July

Explanation:

We all know the sun rises in the east and sets in the west.

No matter where you are on Earth excluding the North and South Poles, you have a due east and due west point on your horizon. That point marks the intersection of your horizon with the celestial equator, the imaginary great circle above the true equator of the Earth.

This is to say that the sun rises close to due east and sets close to due west, for everyone at the equinox. The equinox sun is on the celestial equator. The celestial equator intersects your horizon at due east and due west irrespective of where you are on earth.

The sun rises due east and sets due west during the spring and fall equinoxes. Other times, the sun rises either north or south due east. There are slight changes in the rising and setting of the sun each day. At summer solstice, the sun rises far to the north east and set to the north west. Everyday, the sun rises a bit in furtherance to the south.

Therefore in Newyork state, during summer in July, the sun rise north of due east.

"The correct answer is (2) July.  To understand why the sun rises north of due east in New York State during the month of July, we need to consider the concept of the solar path and the tilt of the Earth's axis.

The Earth orbits the Sun with an axial tilt of approximately 23.5 degrees. This tilt is the reason we experience seasons. As the Earth orbits the Sun, the direction of sunrise and sunset changes throughout the year for any given location north or south of the tropics.

In the Northern Hemisphere, the sun rises exactly due east on two days of the year: the spring (vernal) equinox and the autumnal equinox. These occur around March 20th and September 22nd, respectively. On these days, the sunrise location is at its southernmost point along the horizon for the year.

After the spring equinox, the sunrise location begins to move northward along the horizon each day, reaching its northernmost point on the summer solstice, which occurs around June 21st. After the summer solstice, the sunrise location starts to move southward again, but it continues to rise north of due east until several weeks after the solstice.

For New York State, which is located in the Northern Hemisphere at a latitude of approximately 43 degrees north, the sun rises north of due east from late March until late September. July is well within this period, confirming that the sun rises north of due east during this month.

By contrast, in February, the sun is still moving northward towards the due east position, and by October, it has already passed its northernmost rising point and is moving back towards due east. In December, the sun rises well south of due east.

Therefore, the correct answer to the question is July, when the sun rises north of due east in New York State."

Answer:

v = 196 - 186*e^( - 0.05*t )      

v-terminal = 196 m/s

Explanation:

Given:

- The differential equation for falling object velocity v in gravity with air resistance is given by:

                                m*dv/dt = m*g - b*v

- The initial conditions and constants are as follows:

                 v(0) = 10 , m = 100 kg , b = 5 kg/s , g = 9.8 m/s^2  

Find:

- Find a formula for the velocity of the object at time t. Further, find the terminal (or limiting) velocity of the object. Circle your velocity formula and the terminal velocity.

Solution:

- Rewrite the differential equation in te form:

                                 dv/dt + (b/m)*v = g

- The integration factor function P(t) = b/m. The integrating factor u(t) is:

                                 u(t) = e^∫P(t).dt

                                 u(t) = e^∫(b/m).dt

                                 u(t) = e^[(b/m).t]

- Solve the differential equation after expressing in form:

                                 v.u(t) = ∫u(t).g.dt    

                                 v.e^[(b/m).t] = g*∫e^[(b/m).t].dt    

                                 v.e^[(b/m).t] = g*m*e^[(b/m).t] / b + C

                                  v = g*m/b + C*e^[-(b/m).t]                

- Apply the initial conditions v(0) = 10 m/s and evaluate C:

                                  10 = 9.8*100/5 + C*e^[-(b/m).0]

                                  10 = 9.8*100/5 + C

                                  C = -186

- The final ODE solution is:

                                  v = 196 - 186*e^( - 0.05*t )

- The Terminal velocity vt can be expressed by a limiting value for v(t), where t ->∞.        

                                  vt = Lim t ->∞ ( v(t) )

                                  vt = Lim t ->∞ ( 196 - 186*e^( - 0.05*t ) )

                                  vt = 196 - 0 = 196 m/s

Answer:

power developed by the turbine = 18342 kW  

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 =   2400 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 30 kg/s

solution

first we apply here  thermodynamic equation that is express as energy equation is  

[tex]h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w[/tex]      .......................1

we know turbine is insulated so q is  0

put here value we get

[tex]3015.4 \times 1000 + \frac{10^2}{2} = 2400 \times 1000 + \frac{90^2}{2} + w[/tex]    

w =  611400 J/kg = 611.4 kJ/kg  

and  now we get power developed by the turbine W is

W = mass flow rate × w     ................2

put here value

W = 30 × 611.4

W = 18342 kW  

power developed by the turbine = 18342 kW  

Explanation:

Below is an attachment containing the solution.

To calculate the current, we first convert the mass of gold deposited to moles and then use the equivalence of one mole of gold requiring one faraday of charge. We calculate the total charge transferred and then find the average current by dividing this charge by the total time in seconds.

To find the current in the cell during the period when 4.00 × 10⁻³ kg of gold was deposited on the negative electrode, we need to calculate the total charge that resulted in the deposition of gold. First, we will convert the mass of gold to moles using the molar mass of gold, which is approximately 197 g/mol. Since gold ions carry one unit of positive charge, one mole of gold ions will require one faraday (96,485 C) to be reduced and deposited as gold metal.

After calculating the number of moles of gold, we can then determine the number of moles of electrons that moved through the cell using the equivalence of moles of electrons to moles of gold deposited. Multiplying this number by the charge per mole of electrons (1 faraday), we get the total charge moved. To find the current, we divide the total charge by the total time in seconds, and this gives us the average current over the 2.73 hours.

Answer:

a) When the current is from west to east and the magnetic field is from south to north the magnitude of the force is 2.1x10⁻⁴N and the direction is upwards.

b) The current is moving vertically upward, the magnitude of the force is 2.1x10⁻⁴N and the direction is west.

c) The force is 0 because the magnetic field and the direction of the current are in parallel.

d) No, the force is less.

Explanation:

Given:

L = length of the wire = 2.5 m

i = current in wire = 1.5 A

B = magnetic field = 0.55x10⁻⁴T

a) The magnitude of magnetic force is equal to:

[tex]F=BiL=0.55x10^{-5} *1.5*2.5=2.1x10^{-4} N[/tex]

b) The same way to a):

F = 2.1x10⁻⁴N

c) F = 0

The magnetic field and the direction of the current are in parallel.

d) The answer is no, the force is less

Answer:

the case is the one  with the greatest current, L=15 cm ,   i = 2.19 10⁸  A

Explanation:

Ohm's law is

          V = i R

Resistance is

         R = ρ L / A

Where L is the length of the electrons pass and A the area perpendicular to the current

      i = V / R

      i = V (A / ρ L)

      i = V / ρ  (A / L)

We can calculate the relationship between the area and the length to know in which direction the maximum currents

Case 1

      L = 0.15 m

      A = 0.26 0.43 = 0.1118 m2

      A / L = 0.1118 / 0.15

      A / L = 0.7453 m

Case 2

        L = 0.26 m

        A = 0.15 0.43 = 0.0645 m2

        A / L = 0.248 m

Case 3

       L = 0.43 m

       A = 0.15 0.26 = 0.039 m2

        A / L = 0.0907 m

We can see that the case is the one  with the greatest current, L=15 cm

Let's calculate the current

     i = 5 / 1.7 10⁻⁸ (0.7453)

      i = 2.19 10⁸  A