Answer:
e. 1.2 x 10²³
Explanation:
According to the problem, The current equation is given by:
[tex]I(t)=0.88e^{-t/6\times3600s}[/tex]
Here time is in seconds.
Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.
The relation between current and number of charge carriers is:
[tex]q=\int\limits {I} \, dt[/tex]
Here the limits of integration is from 0 to infinite. So,
[tex]q=\int\limits {0.88e^{-t/6\times3600s}}\, dt[/tex]
[tex]q=0.88\times(-6\times3600)(0-1)[/tex]
q = 1.90 x 10⁴ C
Consider N be the total number of charge carriers. So,
q = N e
Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.
N = q/e
Substitute the suitable values in the above equation.
[tex]N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}[/tex]
N = 1.2 x 10²³
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed on a screen 2.60 m away. Define the width of a bright fringe as the distance between the minima on either side.
Answer:
[tex]W = 10.28\ mm[/tex]
Explanation:
Given,
Red light wavelength = 633 nm
width of slit = 0.320 mm
distance,d = 2.60 m
Condition of first maximum
[tex]a sin \theta_1 = m\lambda [/tex]
[tex]\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})[/tex]
m = 1
[tex]\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})[/tex]
[tex]\theta_1 = 0.1133^\circ[/tex]
Width of the first minima
[tex]y_1 = L tan \theta_1[/tex]
[tex]y_1 = 2.60\times tan( 0.11331)[/tex]
[tex]y_1 = 5.14 \ mm[/tex]
Now, width of the central region
[tex]W = 2 y_1[/tex]
[tex]W = 2\times 5.14[/tex]
[tex]W = 10.28\ mm[/tex]
If the frequency of the radio station is 88.1MHz(8.81 •10^7Hz), what is the wavelength of the wave used by the radio station for its broadcast? The answer should have three significant figures
Answer:
3.41m
Explanation:
The following were obtained from the question:
f (frequency) = 8.81x10^7Hz
V (velocity of electromagnetic wave) = 3x10^8 m/s
λ (wavelength) =?
Velocity, frequency and wavelength of a wave are related with the equation below:
V = λf
λ = V/f
λ = 3x10^8 /8.81x10^7
λ = 3.41m
Therefore, the wavelength of the radio wave is 3.41m
Answer:
Answer: 3.41
Explanation:
Edge 2020 (E2020)
A particle of mass m is confined to a box of length`. Its initial wave function is identical to that of the displacement of the string in the problem above, Boas Ch. 13, Sec. 4, #4.Find the solution of the Schrodinger equation
Answer:
φ = √2/L sin (kx), E = (h² / 8 mL²) n²
Explanation:
The Schrödinger equation for a particle in a box is, described by a particle within a potential for simplicity with infinite barrier
V (x) = ∞ x <0
0 0 <x <L
∞ x> L
This means that we have a box of length L
We write the equation
(- h’² /2m d² / dx² + V) φ = E φ
h’= h / 2π
The region of interest is inside the box, since being the infinite potential there can be no solutions outside the box. The potential is zero
- h’² /2m d²φ/ dx² = E φ
The solution for this equation is a sine wave,
Because it is easier to work with exponentials, let's use the reaction between the sine function and cook with the exponential
[tex]e^{ikx}[/tex] = cos kx + i sin kx
Let's make derivatives
dφ / dx = ika e^{ikx}
d²φ / dx² = (ik) e^{ikx} = - k² e^{ikx}
Let's replace
- h'² / 2m (-k² e^{ikx}) = E e^{ikx}
E = h'² / 2m k²
To have a solution this expression
Now let's work on the wave function, as it is a second degree differential bond, two solutions must be taken
φ = A e^{ikx} + B e^{-ikx}
This is a wave that moves to the right and the other to the left.
Let's impose border conditions
φ (0) = 0
φ (L) = 0
For being the infinite potential
With the first border condition
0 = A + B
A = -B
They are the second condition
0 = A e^{ikL}+ B e^{-ikL}
We replace
0 = A (e^{ikL} - e^{-ikL})
We multiply and divide by 2i, to use the relationship
sin kx = (e^{ikx} - e^{-ikx}) / i2
0 = A 2i sin kL
Therefore kL = nπ
k = nπ / L
The solution remains
φ = A sin (kx)
E = (h² / 8 mL²) n²
To find the constant A we must normalize the wave function
φ*φ = 1
A² ∫ sin² kx dx = 1
We change the variable
sin² kx = ½ (1 - cos 2kx)
A =√ 2 / L
The definitive function is
φ = √2/L sin (kx)
Three individual point charges are placed at the following positions in the x-y plane:Q3= 5.0 nC at (x, y) = (0,0);Q2= -3.0 nC at (x, y) = (4 cm, 0); and Q1= ?nC at (x, y) = (2 cm,0);What isthe magnitude, and sign, ofcharge Q1such that the net force exerted on charge Q3, exerted bycharges Q1and Q2, is zero?
Answer:
Explanation:
net force exerted on charge Q₃, exerted by charges Q₁and Q₂, will be zero
if net electric field due to charges Q₁ and Q₂ at origin is zero .
electric field due to Q₂
= 9 X 10⁹ X 3 x10⁹ / .04²
electric field due to Q₁
= 9 X 10⁹ X Q₁ / .02²
For equilibrium
9 X 10⁹ X Q₁ / .02² = 9 X 10⁹ X 3 x10⁻⁹ / .04²
Q₁ = 3 X10⁻⁹ x .02² / .04²
= 3 / 4 x 10⁻⁹
.75 x 10⁻⁹ C
A SMA wire in the un-stretched condition is then given an initial strain of εo (to preload the wire) at room temperature (RT). Its ends are then rigidly fixed. What force is developed in the wire?
Answer: tensional force
Explanation:
Tension force on a material occurs when two equal forces act on a material in an opposite direction away from the ends of the material.
Pre-tensing a wire material increases its load bearing capacity and reduces its flexure.
Points A, B, C, and D are at the corners of a square area in an electric field, with B adjacent to A and C diagonally across from A. The potential difference between A and C is the negative of that between A and B and the same as that between B and D. Part B What is the potential difference between C and D? Delta V_CD = Delta V_AB Delta V_CD = -Delta V_AB Delta V_CD = -2 Delta V_AB Delta V_CD = 0 Part C What is the potential difference between A and D? Delta V_CAD = Delta V_AB Delta V_AD = -2 Delta V_AB Delta V_AD = Delta V_AB Delta V_AD = 0
For a square in an electric field, the potential difference between points C and D is the same as that between A and B, and the potential difference between A and D is zero.
Explanation:To answer these questions, we first need to understand the concept of electric potential difference, or voltage. It's defined as the change in potential energy of a charge moved between two points, divided by the charge. In the case of a square area in an electric field, if the potential difference between points A and B (ΔV_AB) and between B and D (ΔV_BD) is x volts, then the potential difference between A and C (ΔV_AC) is -x volts, as it's negative of ΔV_AB. Due to this, the potential difference between C and D (ΔV_CD) would be also x volts as it's same as that between A and B and equal to ΔV_BD.
For Part C of the question, thinking about the square as a cycle, if we traverse from A to B to D (or vice versa) the total potential difference would be x - x = 0 volts. Therefore, the potential difference between A and D (ΔV_AD) is zero.
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An owl is carrying a vole in its talons, flying in a horizontal direction at 8.3 m/s while 282 m above the ground. The vole wiggles free, and it takes the owl 2 s to respond. When it does respond, it dive at a constant speed in a straight line, catching the mouse 2 m from the ground (a) What is the owl's dive speed? (b) What is the owl's dive angle below the horizontal?(n radians) (c) How long, in seconds, does the mouse fall?
a) 37.9 m/s
b) 1.35 rad below horizontal
c) 7.56 s
Explanation:
a-c)
At the beginning, both the owl and the vole are travelling in a horizontal direction at a speed of
[tex]v_x=8.3 m/s[/tex]
After the vole wiggles free, the owl takes 2 seconds to react; the horizontal distance covered by the owl during this time is
[tex]d_x = v_x t =(8.3)(2)=16.6 m[/tex]
The vertical motion of the wiggle is a free fall motion, so it is a uniformly accelerated motion with constant acceleration
[tex]g=9.8 m/s^2[/tex] in the downward direction
The wiggle falls from a height of h' = 282 m to a height of h = 2 m, so the vertical displacement is
s = h' - h = 282 - 2 = 280 m
The time it takes the wiggle to cover this distance is given by the suvat equation:
[tex]s=u_y t - \frac{1}{2}gt^2[/tex]
where [tex]u_y = 0[/tex] is the initial vertical velocity. Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(280)}{9.8}}=7.56 s[/tex]
The owl must cover the vertical distance of 280 m in this time interval, so its vertical speed must be:
[tex]v_y=\frac{s}{t}=\frac{280}{7.56}=37.0 m/s[/tex]
Therefore, the speed of the owl during the dive is the resultant of the velocities in the two directions:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{8.3^2+37.0^2}=37.9 m/s[/tex]
b)
In part a-c, we calculated that the components of the velocity of the owl in the horizontal and vertical direction, and they are
[tex]v_x=8.3 m/s\\v_y=37.0 m/s[/tex]
This means that the angle of the owl's dive, with respect to the original horizontal direction, is
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]
And substituting these values, we find:
[tex]\theta=tan^{-1}(\frac{37.0}{8.3})=77.4^{\circ}[/tex]
And this angle is below the horizontal direction.
Converting into radians,
[tex]\theta=77.4\cdot \frac{2\pi}{360}=1.35 rad[/tex]
Final answer:
The owl's dive speed is 7.15 m/s, dive angle is 30 degrees below the horizontal, and it takes about 1.76 seconds for the mouse to fall.
Explanation:
The owl's dive speed: Using the equation of motion for constant acceleration, we can find that the owl's dive speed is 7.15 m/s.
The owl's dive angle below the horizontal: The owl's dive angle is 30 degrees above the horizontal when diving.
How long the mouse falls: Applying the kinematic equation for vertical motion, the time it takes for the mouse to fall completely is approximately 1.76 seconds.
Suppose the dim-looking headlight on the right is actually a small light on the front of a bicycle. What can you conclude about the distance of the motorcycle and bicycle?
Answer:
Explanation:
The dimness of two light source, which have the same intensify depends on their distance from the observer. That is why some stars appear brighter than the others, this is due to the distance of each of them from the earth.
The closer a light source, the brighter they appear
Now, if the right lamp has a dim light actually, it shows that the bicycle is closer than it appears.
Two identical black holes collide head-on. Each of them has a mass equivalent to 37 solar masses. (The sun has a mass of about 2×1030 kg.) As the black holes collide, they merge, forming a single, larger black hole and additional gravitational waves that carry momentum out of the system. Before the collision, one black hole is moving with a speed of 56 km/s, while the other one is moving at 69 km/s. After the collision the larger black hole moves with speed 4 km/s. How much momentum was carried away by gravitational waves?
Answer:
[tex]3.7\times10^{35}\text{ kg m/s}[/tex]
Explanation:
The system of the colliding bodies is ideally isolated, so no external forces act on it. By the principle of conservation of linear momentum, the total initial momentum is equal to the total final momentum.
Both bodies had a head-on collision. We take the direction of the faster body as the positive direction. Because they have the same mass, let's call this mass m.
Hence, we have for the initial momentum
[tex]69m - 56m = 13m[/tex]
The final momentum is
[tex](m+m) \times4 =[/tex]8m
The difference in both momenta is the momentum carried by the gravitational waves.
[tex]13 m - 8m = 5m[/tex]
Converting to the appropriate units and using the actual value of m (37 × a solar mass), we have
[tex]5\times10^3 \text{ m/s}\times37\times2\times10^{30} \text{ kg} = 3.7\times10^{35}\text{ kg m/s}[/tex]
What is the voltage across six 1.5-V batteries when they are connected (a) in series, (b) in parallel, (c) three in parallel with one another and this combination wired in series with the remaining three?
The resultant voltage depends on the arrangement of the batteries. For series configuration, the voltage sums up to 9V. For parallel, it remains 1.5V. And for combined series and parallel, it sums up to 3V.
Explanation:The voltage across batteries depends on how they are connected.
When the batteries are connected in series, the voltages add up. So, for six 1.5-V batteries, the total voltage is 6 * 1.5V = 9V. When the batteries are connected in parallel, the voltage remains the same as one battery, which is 1.5V, no matter how many batteries are connected. If three batteries are connected in parallel with each other and then in series with the remaining three also organized in parallel, the voltage would be 1.5V (parallel group) + 1.5V (parallel group) = 3V.Learn more about Voltage here:https://brainly.com/question/31347497
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A truck runs into a pile of sand, moving 0.80 m as it slows to a stop. The magnitude of the work that the sand does on the truck is 5.5×105J. Part A Determine the magnitude of the average force that the sand exerts on the truck
Answer:
687,500 N
Explanation:
Workdone = Force × Distance
Making force the subject of the formula; we have:
Force =[tex]\frac{workdone}{distance}[/tex]
Given that:
workdone = 5.5×10⁵ J
Distance = 0.80 m
∴ Force = [tex]\frac{5.5*10^5}{0.8}[/tex]
Force = 687,500 N
Answer:
6.875×10⁵ N.
Explanation:
Force: This can be defined as the product of mass and acceleration or it can be defined as the ratio of work done and distance. The S.I unit of force is Newton.
W = F×d................. Equation 1
Where W = work done, F = force, d = distance.
make F the subject of the equation
F = W/d.................... Equation 2
Given: W = 5.5×10⁵ J, d = 0.8 m
Substitute into equation 2
F = 5.5×10⁵ /0.8
F = 6.875×10⁵ N.
Hence the force exerted on the truck by the sand = 6.875×10⁵ N.
Jen falls out of a tree and lands on a trampoline. The trampoline sags 60 cm before launching Jen back into the air. At the very bottom, where the sag is the greatest, is Jen’s acceleration upward, downward, or zero?
Answer:
At the very bottom, whnere the sag is the greatest, Jay’s acceleration is upward.
Explanation:
As Jay lands on the trampoline, Jay’s motion was dowward, the trampoline was opposing his motion and hence, caused him to reach an initial halt position. Afterwards, the trampoline causes Jay to move back into the air and therefore, the change in velocity vector act in upward direction. The acceleration vector is always align towards the change in velocity vector's direction.
A thermometer reading 65° F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110° F after 1 2 minute and 140° F after 1 minute. How hot is the oven?
The temperature of the oven is 200°F
Explanation:
Given-
We have to apply Newton's law of cooling or heating
[tex]\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt[/tex]
On Integrating both sides, we get
[tex]T = Tm + Ce^k^t[/tex]
On putting the value,
T(0) = 65°F
[tex]65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t[/tex]
After 1/2 minute, thermometer reads 110°F. So,
[tex]110 = Tm + (65 - Tm)e^0^.^5^k[/tex] - 1
After 1 minute, thermometer reads 140°F. So,
[tex]140 = Tm + (65 - Tm)e^k[/tex] - 2
Dividing equation 2 by 1:
[tex]e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}[/tex] - 3
From 1 we have,
[tex]e^0^.^5^k = \frac{110 - Tm}{65 - Tm}[/tex]
Putting this vale in eqn 3. We get,
[tex]\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}[/tex]
[tex](110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200[/tex]
Therefore, the temperature of the oven is 200°F
The Temperature ( hotness ) of the oven is ; 200°F
Given data :
Initial thermometer reading ( To ) = 65°F
Thermometer reading after 1/2 minute ( T1/2 ) = 110°F
Thermometer reading after 1 minute ( T1 ) = 140°F
Temperature of oven ( Tm ) = ?
Determine the Temperature of the Oven
To determine the temperature of the oven we will apply Newton's law of heating and cooling.
[tex]\frac{dT}{dt} = K(T -Tm )[/tex] ( Integrating the expression we will have )
T = Tm + [tex]Ce^{kt}[/tex] ---- ( 1 )
where k = time
at T = 0 equation becomes
65 = Tm + C
∴ C = 65 - Tm
Back to equation ( 1 )
T = Tm + ( 65 -Tm )[tex]e^{kt}[/tex] ---- ( 2 )
After 1/2 minute equation 2 becomes
110 = Tm + ( 65 - Tm )[tex]e^{0.5k}[/tex] ---- ( 3 )
After 1 minute equation 2 becomes
140 = Tm + ( 65 - Tm )[tex]e^{k}[/tex] ---- ( 4 )
Next step : Divide equation 4 by equation 3
[tex]e^{0.5k} = \frac{140 - Tm}{110 - Tm}[/tex] ----- ( 5 )
Also From equation 3
[tex]e^{0.5k} = \frac{110 - Tm}{65 - Tm}[/tex] --- ( 6 )
Next step : Equating equations ( 5 ) and ( 6 )
( 110 - Tm )² = ( 140 - Tm )( 65 - Tm )
3000 - 220 Tm = -205Tm
∴ Tm = 3000 / 15 = 200°F
Hence we can conclude that the temperature ( hotness ) of the oven is 200°F.
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A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubber square of the same size. The rubber dielectric is 2.5 mm thick, hasr2.5, andbreakdown field strength of 25 megavolts per meter. Find the voltage rating of the capacitor using a safetyfactor of 10.
Answer:
The voltage will be 0.0125V
Explanation:
See the picture attached
he frequency of the stretching vibration of a bond in IR spectroscopy depends on A) the strengthof the bond and the electronegativity of the atomsB) the electronegativity of the atoms and the nuclear charges of the atomsC) the electronegativity of the atoms and the masses of the atomsD)the masses of the atoms and the strengthof the bond
Answer:
D) the masses of the atoms and the strength of the bond.
Explanation:
Mostly in diatomic and triatomic molecules, bonds of the molecules experience vibrations and rotations. When there is a continuous change in the bond distance of the two atoms then these vibrations are termed as stretching vibrations.
As these vibrations exist in the bonds between the atoms. So they depend upon the masses of the atoms and strength of the bonds. Greater masses of the atoms and strong bond strength will result in reduction of vibration. Thats why we don't observe such stretching vibrations in larger, massive molecules. They mostly exists in the diatomic and triatomic molecules where the bond strength is not that much stronger and the masses of the atoms are small.
Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much current flows through any one of the resistors
Answer:
4 A
Explanation:
We are given that
[tex]R_1=R_2=R_3=4\Omega[/tex]
I=12 A
We have to find the current flowing through each resistor.
We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.
Formula :
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]
Using the formula
[tex]\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}[/tex]
[tex]R=\frac{4}{3}\Omega[/tex]
[tex]V=IR[/tex]
Substitute the values
[tex]V=12\times \frac{4}{3}=16 V[/tex]
[tex]I_1=\frac{V}{R_1}=\frac{16}{4}=4 A[/tex]
[tex]I_1=I_2=I_3=4 A[/tex]
Hence, current flows through any one of the resistors is 4 A.
The current flows through a resistor is equal to 4A.
To understand more, check below explanation.
Parallel connection of resistors:When n number of resistors are connected in parallel to a battery and current I flowing in the circuit.
Then, current flows in each resitor [tex]=\frac{I}{n}[/tex]
It is given that, three identical resistors are connected in parallel to a battery and total current of 12A flows through the circuit.
So that, current in each resistor [tex]=\frac{12}{3}=4A[/tex]
Hence, the current flows through a resistor is equal to 4A.
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You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are present for the testing of the ride, in which an empty 220 kg car is sent along the entire ride. Near the end of the ride, the car is at near rest at the top of a 101 m tall track. It then enters a final section, rolling down an undulating hill to ground level. The total length of track for this final section from the top to the ground is 250 m. For the first 230 m, a constant friction force of 350 N acts from computer-controlled brakes. For the last 20 m, which is horizontal at ground level, the computer increases the friction force to a value required for the speed to be reduced to zero just as the car arrives at the point on the track at which the passengers exit. (a) Determine the required constant friction force (in N) for the last 20 m for the empty test car.
Answer:
The required constant friction force for the last 20 m is 6,862.8 N
Explanation:
Energy Conversion
There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.
The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.
Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:
[tex]E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J[/tex]
For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus
[tex]E_1=K_1+W_1[/tex]
We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:
[tex]W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J[/tex]
This means that the kinetic energy that remains when the car reaches ground level is
[tex]K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J[/tex]
We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:
[tex]W_2=F_{r2}\cdot x_2=137,256[/tex]
Solving for Fr2
[tex]\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N[/tex]
The required constant friction force for the last 20 m is 6,862.8 N
To determine the required constant friction force for the last 20 m of the roller coaster ride, we consider the energy changes that occur. The work done by the friction force is equal to the loss in potential energy. By using the equation -mgh = -f x d, we can solve for the friction force.
Explanation:To determine the required constant friction force for the last 20 m of the roller coaster ride, we need to consider the energy changes that occur. As the car rolls downhill, it loses potential energy and gains kinetic energy. The work done by the friction force is equal to the loss in potential energy. Using the equation -m g h = - f x d, where m is the mass of the car, g is the acceleration due to gravity, h is the height of the track, f is the friction force, and d is the distance over which the force acts, we can solve for the friction force.
In this case, the mass of the car is 220 kg, the height of the track is 101 m, and the distance over which the force acts is 20 m. Plugging these values into the equation, we get:
-220 kg * 9.8 m/s² * 101 m = -f * 20 m
Solving for f, we find that the required constant friction force for the last 20 m of the ride is approximately 2151 N.
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Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to these charges at a point at on the x-axis?
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
[tex]V =\frac{Kq}{r}[/tex]
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V[/tex]
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
[tex]V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V[/tex]
Total potential due to this charges = 4500 V + 6750 V = 11250 V
The potential due to two 3.0 μC charges on the x-axis at different distances from a point can be calculated using Coulomb's Law.
Explanation:The potential due to two point charges can be found using Coulomb's Law. The potential, V, at a point on the x-axis is the sum of the potentials from each charge. The potential due to a point charge can be calculated using the formula V = k * (Q / r), where k is the electrostatic constant, 9 x 10^9 Nm^2/C^2, Q is the charge, and r is the distance between the charge and the point. In this case, since the charges are on the x-axis, the distance between the origin and the point is x, and the distance between the other charge and the point is (6-x). So, the potential at the point is V = k * (3.0 x 10^-6 / x) + k * (3.0 x 10^-6 / (6-x)) relative to infinity.
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A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H. L=2.55 H. What is the induced emf 1.43 s 1.43 s after the circuit has been closed?
Answer:
The induced emf 1.43 s after the circuit is closed is 4.19 V
Explanation:
The current equation in LR circuit is :
[tex]I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })[/tex] .....(1)
Here I is current, V is source voltage, R is resistance, L is inductance and t is time.
The induced emf is determine by the equation :
[tex]V_{e}=L\frac{dI}{dt}[/tex]
Differentiating equation (1) with respect to time and put in above equation.
[tex]V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }[/tex]
[tex]V_{e}=Ve^{\frac{-Rt}{L} }[/tex]
Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.
[tex]V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }[/tex]
[tex]V_{e}=4.19\ V[/tex]
Based on the calculations, the induced emf is equal to 4.19 Volts.
Given the following data:
Voltage = 6.05 V.Resistance = 0.655 Ω.Inductance = 2.55 H.Time = 1.43 seconds.How to determine the induced emf.In a RL circuit, current is given by this mathematical expression:
[tex]I=\frac{V}{R} (1-e^{\frac{Rt}{L} })[/tex]
Where:
I is the current.V is the source voltage.R is the resistance.L is the inductance.t is the time.For an induced emf in a circuit, we have:
[tex]E=L\frac{dI}{dt} \\\\E=L \times I \\\\E=L \times \frac{V}{R} (1-e^{\frac{-Rt}{L} })\\\\E=V e^{\frac{-Rt}{L} }\\\\E=6.05 e^{\frac{-0.655 \times 1.43}{2.55} }[/tex]
E = 4.19 Volts.
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Two inclined planes A and B have the same height but different angles of inclination with the horizontal. Inclined plane A has a steeper angle of inclination than inclined plane B. An object is released at rest from the top of each of the inclined planes.
How does the speed of the object at the bottom of inclined plane A compare with that of the speed at the bottom of inclined plane B?
Answer:
It is the same.
Explanation:
Assuming no friction between the object and the surface, and no other external force acting on the object, than gravity and normal force, we can say the following:[tex]\Delta K + \Delta U = 0[/tex]
where ΔK = change in kinetic energy, and ΔU = change in gravitational potential energy.As ΔU = -m*g*h (being h the height of the plane), it will be the same for both inclined planes, as we are told that they have the same height.If the object starts from rest, the change in kinetic energy will be as follows:[tex]\Delta K = K_{f} - K_{0} = \frac{1}{2} * m*v_{f} ^{2} (1)[/tex]
[tex]\Delta K = -\Delta U = m*g*h (2)[/tex]
From (1) and (2) we see that the mass m and the height h are the same, the speed at the bottom of inclined plane A, will be the same as the one at the bottom of inclined plane B.Despite the difference in angle, the speed of the objects at the bottom of the planes will be the same because they start with the same potential energy at the top that's entirely converted to kinetic energy (the energy of motion) by the bottom, so long as energy losses are ignored.
Explanation:The subject of this question lies in the realm of Physics, specifically involving principles of mechanical energy and gravitational potential energy. In the stated scenario, the two objects start on their respective inclined planes from a state of rest. Accordingly, they possess potential energy but no kinetic energy.
As the objects slide down their respective planes, this potential energy is converted into kinetic energy—the energy of motion. Because the two planes are of identical height (thus imparting the same initial potential energy to the objects), and because all potential energy will have been converted to kinetic energy by the time the objects reach the bottom (ignoring energy losses due to friction or air resistance), both objects will possess the same kinetic energy—and thereby the same speed—at the bottom of their planes, regardless of the angle of inclination.
In practical applications, friction and other factors may have a role and might cause the object on the steeper plane (Plane A) to reach the bottom more quickly. However, that's not due to a difference in speed at the bottom; it's about the time taken to get there.
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Air "breaks down" when the electric field strength reaches 3 × 106 N/C, causing a spark. A parallel-plate capacitor is made from two 3.0 cm × 3.0 cm electrodes.How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?
Answer:
[tex]1.5\times 10^{11}}[/tex]
Explanation:
We are given that
Electric field=[tex]E=3\times 10^6 N/C[/tex]
Dimension of parallel plate capacitor=[tex]3 cm\times 3 cm[/tex]
Area of parallel plate capacitor=[tex]A=3\times 3=9 cm^2=9\times 10^{-4}m^2[/tex]
[tex]1 cm^2=10^{-4} m^2[/tex]
We have to find the number of electrons must be transferred from one electrode to the other to create a spark between the electrodes.
[tex]E=\frac{Q}{\epsilon_0A}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]
Substitute the values
[tex]3\times 10^6=\frac{Q}{8.85\times 10^{-12}\times 9\times 10^{-4}}[/tex]
[tex]Q=3\times 10^6\times 8.85\times 10^{-12}\times 9\times 10^{-4}}[/tex]
[tex]Q=2.4\times 10^{-8} C[/tex]
We know that
[tex]Q=ne=n\times 1.6\times 10^{-19} [/tex]
Where e=[tex]1.6\times 10^{-19} C[/tex]
[tex]n=\frac{Q}{e}=\frac{2.4\times 10^{-8}}{1.6\times 10^{-19}}=1.5\times 10^{11}}[/tex]
The driving force for fluid flow is the pressure difference, and a pump operates by raising the pressure of a fluid (by converting the mechanical shaft work to flow energy). A gasoline pump is measured to consume 3.8 kW of electric power when operating, If the pressure differential between-the outlet and inlet of the pump is measured to be 7 kPa and the changes in velocity and elevation are negligible, determine the maximum possible volume flow rate of gasoline.
Answer:
[tex]\dot V = 0.542 \frac{m^{3}}{s}[/tex]
Explanation:
The power needed for the pump to raise the pressure of gasoline is defined by following equation. The maximum possible volume flow rate is isolated and then calculated:
[tex]\dot W = \dot V \cdot \Delta P\\\dot V = \frac{\dot W}{\Delta P}\\\dot V = \frac{3.8 kW}{7 kPa}\\\dot V = 0.542 \frac{m^{3}}{s}[/tex]
Explanation:
Below is an attachment containing the solution.
A solenoid has length L, radius R, and number of turns N. A second smaller solenoid of length L, radius R and number of turns N is placed at the center of the first solenoid, such that their axes coincide. What is the mutual inductance of the pair of solenoids
Answer:
Explanation:
Mutual inductance is FLUX induced in second coil due to unit current passed in first coil .
Let i be the current in the bigger coil.
magnetic field at its center
B = μ₀ n i , n is no of turns per unit length
= μ₀ (N / L) i
Magnetic flux associated with small coil placed near its axis
= B X πR² X N
=μ₀ (N / L) i X πR² X N
FLUX = μ₀ (N² / L) i X πR²
FLUX induced by unit current
M = μ₀ (N² / L) X πR²
Singly charged positive ions are kept on a circular orbit in a cyclotron. The magnetic field inside the cyclotron is 1.833 T. The mass of the ions is 2.00×10-26 kg, and speed of the ions is 2.05 percent of the speed of the light. What is the diameter of the orbit? (The speed of the light is 3.00×108 m/s.)
Answer:
The diameter is 0.8376 m
Explanation:
Magnetic force is the force that is associated with the magnetic field, the magnitude of the magnitude force can be obtained using equation 1;
F = q v B .....................................1,
where q is the magnitude of the charge of the particle =;
v is the velocity and;
B is the magnetic field = 1.833 T ;
Here, the path of the charge is circular so the force can be also considered as the centripetal force which is represented in equation 2;
[tex]F_{c}[/tex] = m [tex]v^{2}[/tex] / r ....................................2,
m is the particle's mass = 2.00 x[tex]10^{-26}[/tex] kg
v is the speed of ion = 2.05% of 3.00 x [tex]10^{8}[/tex] = 0.0205 x 3.00 x [tex]10^{8}[/tex]
= 6.15 x [tex]10^{6}[/tex] m/s ;
Singly charged ion has a charge equal to the electron charge and the magnitude = 1.60217646 ×[tex]10^{-19}[/tex] C and;
r is the radius of the circular path.
to get the diameter of the orbit we equate equation 1 to 2 and isolate r in equation 2.
q v B = m [tex]v^{2}[/tex] / r
r = m v/q B.....................................3
r = (2.00 x[tex]10^{-26}[/tex] kg) x (6.15 x [tex]10^{6}[/tex] m/s) / (1.60217646 ×[tex]10^{-19}[/tex] C) x (1.833 T)
r = 0.4188 m
The diameter is r x 2
D = 0.4188 m x 2
D = 0.8376 m
Therefore the diameter is 0.8376
You accidentally drop a quarter into the hot coals of a campfire. You fish out the hot quarter with a pair of pliers and drop the quarter directly on top of a large 2 kg block of ice to cool it down. In what direction does heat flow:
There is no heat flow.
From the quarter to the block of ice
Answer:
from the quarter to the block of ice
Explanation:
Heat flows from higher temperature to lower temperature until temperature of both bodies are in Equilibrium .
Since block of ice has lower temperature than that of quarter. heat transfer will take from quarter to ice until both have same temperature(in other words temperature are in Equilibrium for quarter and ice)
Final answer:
Heat will flow from the hot quarter to the block of ice as heat always transfers from a warmer object to a cooler one. The ice absorbs the heat and may melt without increasing in temperature due to the phase change.
Explanation:
When a hot quarter is dropped onto a large block of ice, heat will flow from the quarter to the block of ice. This is because heat always flows spontaneously from a hotter object to a cooler one according to thermodynamics. In this case, the hot quarter would lose heat, and the block of ice would absorb it. Even as the ice absorbs heat, it may not increase in temperature as it may be undergoing a phase change from solid to liquid at 0°C. The heat is used to break the bonds between water molecules during the melting process, thus increasing the internal potential energy instead of increasing the kinetic energy, which would raise the temperature.
If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Answer:
[tex]\frac{Q}{Q_0}=1[/tex]
Explanation:
Capacitance is defined as the charge divided in voltage.
[tex]C=\frac{Q}{V}(1)[/tex]
Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:
[tex]V=\frac{V_0}{k}[/tex]
Where k is the dielectric constant and [tex]V_0[/tex] the voltage of the capacitor without a dielectric
The capacitance with a dielectric between the capacitor plates is given by:
[tex]C=kC_0[/tex]
Where k is the dielectric constant and [tex]C_0[/tex] the capacitance of the capacitor without a dielectric. So, we have:
[tex]Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1[/tex]
Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.
The average life span in the United States is about 70 years. Does this mean that it is impossible for an average person to travel a distance greater than 70 light years away from the Earth? (A light year is the distance that light travels in a year.) This is not a yes or no answer. Explain your reasoning.
Answer:
for the people of the Earth traveling they last much more than 70 years
Explanation:
In order to answer this answer we must place ourselves in the context of special relativity, which are the expressions for time and displacement since the speed of light has a finite speed that is the same for all observers.
The life time of the person is 70 years in a fixed reference system in the person this time we will call their own time (t₀), when the person is placed in a ship that moves at high speed, very close to the speed of the light the time or that an observer measures on Earth, the expression for this time is
t = t₀ 1 / √(1 - (v / c)²)
We see that if the speed of the ship is very close to the speed of light the
the value of the root of the denominator is very high, for which for the person on Earth it measures a very large time even when the person on the ship travels has a time within its 70 years of life
In concussion, for the people of the Earth traveling they last much more than 70 years
a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block?(unit=kg) IM GIVING 30 POINTS FOR THE CORRECT ANSWER
Answer:
577g
Explanation:
Given parameters:
Temperature change = 5.9°C
Amount of heat lost = 427J
Unknown:
Mass of the block = ?
Solution:
The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.
H = m c Ф
H is the heat capacity
m is the mass of the block
c is the specific heat capacity
Ф is the temperature change
Specific heat capacity of lead is 0.126J/g°C
m = H / m Ф
m = [tex]\frac{427}{0.126 x 5.9}[/tex] = 577g
Mass of the lead block is 577g
Answer: 0.5654
Explanation:
block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is parallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0
Answer:
Please find attached
Explanation:
The acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
The normal force on each block is calculated as follows;
[tex]F_n_ A = mgcos \theta\\\\F_n_ B = m_ B g[/tex]
The frictional force on block A is calculated as;
[tex]F_f = \mu_k F_n\\\\F_f = \mu_ kg mgcos \theta[/tex]
The horizontal force on block A is given as;
[tex]F_x = mgsin\theta[/tex]
The tension on the string due to each block is given as;
[tex]T_ A = m_ A a\\\\T_ B = m_ B a[/tex]
The net force on the block B is calculated as;
[tex]m_Bg - (T_A + m_Agsin\theta + \mu mgcos\theta) = T_B\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta= T_B + T_ A\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta = a(m_ B+ m_ A)\\\\a = \frac{m_Bg - m_Agsin\theta - \mu mgcos\theta}{m_B + m_ A} \\\\a = \frac{(8)(9.8)\ -\ (10)(9.8)(sin30)\ -\ (0.2)(10)(9.8)(cos30)}{8 + 10} \\\\a = 0.69 \ m/s^2 \ (in -direction \ of \ block \ B)[/tex]
Thus, the acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
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Two identical balls are thrown from the top of a building with the same speed. Ball 1 is thrown horizontally, while ball 2 is thrown at an angle above the horizontal. Neglecting air resistance, which ball will have the greatest speed when hitting the ground below?
Answer:
They both have the same speed when hitting the ground below
Explanation:
Conservation of the Mechanical Energy
In the absence of external non-conservative forces, the total mechanical energy of a particle is conserved or is kept constant.
The mechanical energy is the sum of the potential gravitational and kinetic energies, i.e.
[tex]\displaystyle M=mgh+\frac{mv^2}{2}[/tex]
When the first ball is launched at the top of the building of height h, at a speed vo, the mechanical energy is
[tex]\displaystyle M=mgh+\frac{mv_o^2}{2}[/tex]
When the ball reaches ground level (h=0) the mechanical energy is
[tex]\displaystyle M'=\frac{mv_f^2}{2}[/tex]
Equating M=M'
[tex]\displaystyle mgh+\frac{mv_o^2}{2}=\frac{mv_f^2}{2}[/tex]
We could solve the above equation for vf but it's not necessary because we have derived this relation regardless of the direction of the initial speed. It doesn't matter if it's launched with an angle above the horizontal, directly horizontal, or even directly downwards, the final speed is always the same.
It can also be proven with the exclusive use of the kinematic equations.
Note: The speed is the same for both balls, but not the velocity since the direction of the final velocity will be different in each case. Its magnitude is the same for all cases.
Answer: They both have the same speed when hitting the ground below