Answer:
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
Explanation:
While the satellite is in the space vehicle, it has the next potential energy
[tex]U = -\frac{GmMe}{r}[/tex]
where G is the gravitational constant
m is the satellite's mass in kilograms
Me is the earth's mass
r is the orbit's radius from to the earth's center in meters
[tex]U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}[/tex]
[tex]U = -2.2423x10^{12} J[/tex]
The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi
[tex]U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}[/tex]
[tex]U = -29696124610.3 J[/tex]
Then
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
A ball is tossed vertically upward from a height of 2 m above the ground with an initial velocity of 10 m/s. What will the velocity of the ball be just before it hits the ground?
The velocity of the ball just before it hits the ground, thrown upward from 2 meters with an initial velocity of 10 m/s, can be calculated using kinematic equations and is approximately 11.8 m/s.
To determine the velocity of the ball just before it hits the ground, we can use kinematic equations. One equation that relates initial velocity, acceleration due to gravity, and final velocity is:
v² = u² + 2as
Where:
v is the final velocityu is the initial velocity (10 m/s)a is the acceleration due to gravity (9.8 m/s²)s is the displacement (the initial height plus the distance fallen)In this case, the ball is tossed upward so we consider the acceleration due to gravity as negative (-9.8 m/s²). The displacement, in this case, is 2 m (the height from which the ball is thrown).
Insert the known values into the equation:
v² = (10 m/s)² + 2 × (-9.8 m/s²) × (-2 m)
The negative signs cancel for the displacement and acceleration due to the upward throw and the downward acceleration, which will give us a positive number.
v² = 100 + (19.6 × 2)
v² = 100 + 39.2
v² = 139.2
v = √139.2
v ≈ 11.8 m/s
Therefore, the velocity of the ball just before it hits the ground is approximately 11.8 m/s.
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). At this distance, the free-fall acceleration is g/4.
(a) What is the satellite's orbital speed (m/s)?
(b) What is the period of revolution (min)?
(c) What is the gravitational force on the satellite (N) ?
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.
Orbital Motion(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:
https://brainly.com/question/22247460
The Doppler Effect. In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm. Suppose you see this line at a wavelength of 120.5 nm in star A, at 121.2 nm in Star B, at 121.9 nm in Star C, and at 122.9 nm in Star D. Which stars are coming toward us? Which are moving away? Which star is moving fastest relative to us? What are the speeds of Star B and Star C?
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.
A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experienced by a pilot flying in a circle of constant radius at a constant speed of 475 m / s if the radius of the circle is 1790 m
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
A mass is hanging on the end of a spring and oscillating up and down. There are three forces acting on the mass, the force of the spring, the force of gravity and the force of air resistance. Which of these forces can be associated with potential energy function? a. The spring force only. b. Gravity only. c. Air resistance only. d. The spring force and gravity. e. The spring force and air resistance. f. Gravity and air resistance. g. All three forces.
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
An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly toward her from the back of the stage. The bat will be outfitted with a microphone to pick up the singer's voice and a loudspeaker to rebroadcast the sound toward the audience. The composer wants the sound the audience hears from the bat to be, in musical terms, one half-step higher in frequency than the note they are hearing from the singer. Two notes a half-step apart have a frequency ratio of 2⁽¹/¹²⁾ = 1.059.
With what speed must the bat fly toward the singer?
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
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E= 3.40×105 V/m . When the space is filled with dielectric, the electric field is E= 2.20×105 V/m . Part A What is the charge density on each surface of the dielectric?
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]
A rifle is fired in a valley with parallel vertical walls. The echo from one wall is heard 1.55 s after the rifle was fired. The echo from the other wall is heard 2.5 s after the first echo. How wide is the valley? The velocity of sound is 343 m/s. Answer in units of m.
Answer:
[tex]w=694.575\ m[/tex]
Explanation:
Given:
velocity of the sound, [tex]v=343\ m.s^{-1}[/tex]
time lag in echo form one wall of the valley, [tex]t= 1.55\ s[/tex]
time lag in echo form the other wall of the valley, [tex]t'=2.5\ s[/tex]
distance travelled by the sound in the first case:
[tex]d=v.t[/tex]
[tex]d=343\times 1.55[/tex]
[tex]d=531.65\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s=\frac{d}{2}[/tex]
[tex]s=\frac{531.65}{2}[/tex]
[tex]s=265.825\ m[/tex]
distance travelled by the sound in the second case:
[tex]d'=v.t'[/tex]
[tex]d'=343\times 2.5[/tex]
[tex]d'=857.5\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s'=\frac{d'}{2}[/tex]
[tex]s'=\frac{857.5}{2}[/tex]
[tex]s'=428.75\ m[/tex]
Now the width of valley:
[tex]w=s+s'[/tex]
[tex]w=265.825+428.75[/tex]
[tex]w=694.575\ m[/tex]
To determine the width of the valley with vertical walls using echo timings.
To find the width of the valley:
Calculate the distance to the first wall using the time taken for the first echo.Then, calculate the distance to the second wall using the time taken for the second echo.The width of the valley is the difference between the two distances.Calculations:
Distance to first wall = 343 m/s * 1.55 s = 531.85 m
Distance to second wall = 343 m/s * 2.5 s = 857.5 m
Width of the valley: 857.5 m - 531.85 m = 325.65 m
You throw a rock straight up from the edge of a cliff. It leaves your hand at time t = 0 moving at 13.0 m/s. Air resistance can be neglected. Find both times at which the rock is 4.00 m above where it left your hand. Enter your answers in ascending order separated by a comma. Express your answer in seconds.
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.
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?
1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)
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
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 68.0 m in diameter and has a 0.37 A beam, how many electrons are in the beam
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.
Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands to the turbine exit where the pressure is 0.07 MPa, specific enthalpy is 2400 kJ/kg, and the velocity is 90 m/s. The mass flow rate is 30 kg/s. Neglecting potential energy effects, determine the power developed by the turbine, in kW.
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
An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be _____
(A) three times as large as the initial value.
(B) less than three times as large as the initial value.
(C) more than three times as large as the initial value.
(D) equal to the initial value.
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".
If 4.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.73 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.
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.
Explanation: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.
What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse voltage applied. b. The ideal diode acts as an open circuit regardless of forward voltage or reverse voltage applied. c. The ideal diode acts as a short circuit regardless of forward voltage or reverse voltage applied. d. The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applie
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.
A block slides down a rough ramp (with friction) of height h . Its initial speed is zero. Its final speeds at the bottom of the ramp is v . Choose the system to be the block and the Earth.
While the block is descending, its kinetic energy:
a. Increases.
b. Decreases.
c. Remains constant.
Answer:
a. Increases
Explanation:
Conceptually, when a block of certain mass slides on the rough ramp kept on the earth which happens under the influence of gravity.The block is initially at rest but as the acceleration due to gravity acts on the block at the ramp. The ramp is rough so it applies kinetic friction on the moving block as the block slides down the slope.By the definition we know that the acceleration is the rate of change in velocity and here we have acceleration component in the direction of motion of the block.
Mathematically:
when the body is moving down:
[tex]v=u+gt[/tex]
where:
[tex]v=[/tex] final velocity of the block
[tex]u=[/tex] initial velocity of the block
[tex]g=[/tex] acceleration due to gravity
[tex]t=[/tex] time of observation during the instance of motion
From above it is clear that the velocity of the block will increase as the time passes during the motion.As we know that kinetic energy is given as:
[tex]KE=\frac{1}{2} \times m.v^2[/tex]
where:
[tex]m=[/tex] mass of the block which remains constant (macroscopically)
[tex]v=[/tex] velocity of the block (which increases here as the body descends)
A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field
the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].
The magnetic torque [tex]\( \tau \)[/tex] on a current-carrying loop in a uniform magnetic field can be calculated using the formula:
[tex]\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \][/tex]
Where:
- N is the number of turns in the loop (1 in this case, as there is a single loop),
- I is the current flowing through the loop (2.0 A),
- A is the area of the loop (πr² for a circular loop, where \( r \) is the radius),
- B is the magnetic field strength (0.30 T),
- [tex]\( \theta \)[/tex] is the angle between the normal to the loop's plane and the magnetic field (90° in this case, as the loop is perpendicular to the magnetic field).
Substituting the given values:
[tex]\[ \tau = (1) \cdot (2.0 \, \text{A}) \cdot (\pi \times (0.50 \, \text{m})^2) \cdot (0.30 \, \text{T}) \cdot \sin(90°) \][/tex]
[tex]\[ \tau = 2.0 \cdot \pi \cdot (0.50)^2 \cdot 0.30 \][/tex]
[tex]\[ \tau = 2.0 \cdot \pi \cdot 0.25 \cdot 0.30 \][/tex]
[tex]\[ \tau = 0.15 \cdot \pi \][/tex]
[tex]\[ \tau \approx 0.47 \, \text{N} \cdot \text{m} \][/tex]
So, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is approximately [tex]\( 0.47 \, \text{N} \cdot \text{m} \)[/tex].
Therefore, the correct answer choice is [tex]\(\boxed{\text{0.47 N*m}}\)[/tex].
The complete Question is given below:
A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field?
Answer choices.
.41N*m
.00N*m
.47N*m
.58N*m
.52N*m
The sound intensity at a distance of 16 m from a noisy generator is measured to be 0.25 W/m2. What is the sound intensity at a distance of 28 m from the generator?
Answer:
0.1111 W/m²
Explanation:
If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,
I ∝ (1/r²)
I = k/r²
Since k can be the constant of proportionality. k = Ir²
We can write this relation as
I₁ × r₁² = I₂ × r₂²
I₁ = 0.25 W/m²
r₁ = 16 m
I₂ = ?
r₂ = 24 m
0.25 × 16² = I₂ × 24²
I₂ = (0.25 × 16²)/24²
I₂ = 0.1111 W/m²
The sound intensity at a distance of 28 m from the generator will be "0.1111 W/m²".
Distance and Sound intensityAccording to the question,
Distance, r₁ = 16 m
r₂ = 28 m
Sound intensity, I₁ = 0.25 W/m²
We know the relation,
→ I ∝ ([tex]\frac{1}{r^2}[/tex])
or,
I ∝ [tex]\frac{k}{r^2}[/tex]
Now,
→ I₁ × r₁² = I₂ × r₂²
By substituting the values,
0.25 × (16)² = I₂ × (24)²
I₂ = [tex]\frac{0.25\timers (16)^2}{(24)^2}[/tex]
= 0.1111 W/m²
Thus the above answer is correct.
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A rectangular block of copper has sides of length 15 cm, 26 cm, and 43 cm. If the block is connected to a 5.0 V source across two of its opposite faces, find the following. (a) What is the maximum current the block can carry?
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
During which month does the sun rise north of due east in New York State?
(1) February (3) October
(2) July (4) December
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."
From the top of a cliff overlooking a lake, a person throws two stones. The two stones have identical initial speeds of v0 = 13.4 m/s and are thrown at an angle θ = 30.1°, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?
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.
Explanation: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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A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V line, how large is the back emf?
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.
When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coefficient of volume expansion? PLEASE HELPPP MEEEEE
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:
In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased?
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.
The goal of this experiment is to investigate quantitatively how the magnetic force that a bar magnet and a current-carrying coil exert on each other depends on the distance between them.
Available equipment: Bar magnet, coil with connecting wires, power supply, digital balance, ring stand, clamp, meter stick, tape
Can you design an experiment with that info.
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
Only a small amount of the energy used in an incandescent light bulb (regular bulbs commonly used in households) is actually converted into light. What happens to the rest of the energy
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.
Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass). (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load
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
Jeff was riding in a car. Which change demonstrates the GREATEST effect of inertia?
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.
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ= 12.0°, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was Vo= 18.0 m/s.
With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)?
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.
Assuming that only air resistance and gravity act on a falling object, we can find that the velocity of the object, v, must obey the differential equation dv m mg bv dt . Here, m is the mass of the object, g is the acceleration due to gravity, and b > 0 is a constant. Consider an object that has a mass of 100 kilograms and an initial velocity of 10 m/sec (that is, v(0) = 10). If we take g to be 9.8 m/sec2 and b to be 5 kg/sec, 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.
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