The induced electromotive force (EMF) in a generator coil is directly proportional to the area of the coil, frequency of rotation, and the magnetic field strength, but not to the coil's internal resistance. Changing any of these factors as described, except resistance, results in a proportional change in the induced EMF.
Explanation:Understanding the Factors Affecting Induced EMF in a Generator CoilThe behavior of induced electromotive force (EMF) in a generator is explained by Faraday's law of electromagnetic induction, which states that the induced EMF in a coil is proportional to the rate of change of magnetic flux through the coil.
The average induced EMF doubles if the area A is doubled: True. By doubling the area A, the magnetic flux through the coil also doubles, leading to a doubling of the induced EMF.The average induced EMF doubles if the frequency f is doubled: True. Increasing the frequency of rotation leads to a faster rate of change of magnetic flux, thus doubling the induced EMF.The maximum induced EMF occurs when the coil is rotated about an axis perpendicular to area A: True. This orientation results in the maximum change in magnetic flux, thus inducing the maximum EMF.The average induced EMF doubles if the resistance R is doubled: False. Internal resistance does not affect the induced EMF, which is solely dependent on the change in magnetic flux.The average induced EMF doubles if the magnetic field B is doubled: True. Doubling the magnetic field strength doubles the magnetic flux, which leads to a doubling of the induced EMF.These principles help us to understand the operation of an electric generator, where mechanical work is converted into electrical energy.
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o practice Problem-Solving Strategy 39.1: Particles and Waves. A molecule of hydrogen gas has a mass of 3.35×10−27kg3.35×10−27kg and a diameter of 1.48×10−10m1.48×10−10m. What is the kinetic energy at which this molecule's de Broglie wavelength will be equal to its diameter
Answer:
0.0187 eV
Explanation:
Given that:
diameter of the hydrogen gas (λ) = 1.48 ×10⁻¹⁰ m'
mass of the hydrogen gas = 3.35 ×10⁻²⁷ kg
We need to determine the momentum first before calculating the kinetic energy.
So momentum of the hydrogen gas molecule is written as;
[tex]p =\frac{h}{ \lambda}[/tex]
[tex]p = \frac{96.626*10^{-34}J.s}{1.48*10^{-10}m}[/tex]
[tex]p=4.477*10^{-24} kg.m/s[/tex]
NOW, the kinetic energy of the hydrogen gas molecule is calculated as follows by using the formula:
[tex]k_o = \frac{p^2}{2m}[/tex]
[tex]k_o =\frac{(4.477*10^{-24}kg.m/s)^2}{2(3.35*10^-{27})kg}[/tex]
[tex]k_o=2.9916*10^{-21}J(\frac{1eV}{1.6*10^{-19}J})[/tex]
[tex]K_o=0.0187eV[/tex]
A 0.140 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s . It has a head-on collision with a 0.297 kg glider that is moving to the left with a speed of 2.25 m/s . Suppose the collision is elastic.
Answer:
(a). The final velocity of the first glider is −2.15 m/s.
(b). The final velocity of the second glider is -1.67 m/s.
Explanation:
Given that,
Mass of glider = 0.140 kg
Speed = 0.900 m/s
Mass of another glider = 0.297 kg
Speed =2 .25 m/s
Suppose, Find the magnitude of the final velocity of the first glider.
Find the magnitude of the final velocity of the second glider.
(a). We need to calculate the final velocity of the first glider
Using formula of collision
[tex]v_{1}=\dfrac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}[/tex]
Put the value into the formula
[tex]v_{1}=\dfrac{0.900(0.140+0.297)+2\times(-0.297)\times2.25}{0.140+0.297}[/tex]
[tex]v_{1}=-2.15\ m/s[/tex]
(b). We need to calculate the final velocity of the second glider
Using formula of collision
[tex]v_{2}=\dfrac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}[/tex]
Put the value into the formula
[tex]v_{1}=\dfrac{-2.25(0.297+0.140)+2\times(0.140)\times0.900}{0.140+0.297}[/tex]
[tex]v_{1}=-1.67\ m/s[/tex]
Hence, (a). The final velocity of the first glider is −2.15 m/s.
(b). The final velocity of the second glider is -1.67 m/s.
Answer with Explanation:
We are given that
Mass of glider=m=0.14 kg
Initial speed of first glider, u=0.9 m/s
m'=0.297 kg
Initial speed of second glider, u'=-2.25m/s
a.We have to find the magnitude of final velocity of 0.140 kg glider.
The collision is elastic then the final velocity of 0.140 kg glider
[tex]v=\frac{(m-m')u+2m'u'}{m+m'}[/tex]
Substitute the values
[tex]v=\frac{(0.140-0.297)\times 0.9+2\times 0.297\times (-2.25)}{0.140+0.297}=-3.38m/s[/tex]
Magnitude of final velocity of 0.140 kg glider=3.38 m/s
b.[tex]v'=\frac{u'(m'-m)+2mu}{m+m'}[/tex]
[tex]v'=\frac{-2.25(0.297-0.140)+2\times 0.140\times 0.9}{0.140+0.297}=-0.23 m/s[/tex]
Hence, the magnitude of final velocity of 0.297 kg glider=0.23 m/s
Bulb ܣ is rated for 20W at 12V. Bulb ܤ is rated for 20W at 120V. The two bulbs are mistakenly mixed up so that bulb ܣ is connected to a 120V line and bulb ܤ is connected to a 12V line. a. Determine the power dissipated by Bulb ܣ while connected to the 120Vline. b. Determine the power dissipated by Bulb ܤ while connected to the 12Vline. c. Which bulb, if any, is more likely to burn out and why?
Answer:
a. 2000W
b.0.2W
c. Bulb 1
Explanation:
Data given:
bulb 1 rating =20w,12v
bulb 2 rating =20w,120v.
we calculate the resistance for each bulb
[tex]bulb 1: R=\frac{V^2}{P}\\ bulb 1: R=\frac{12^2}{20}\\ R=7.2 ohms[/tex]
for bulb 2
[tex]bulb 2: R=\frac{V^2}{P}\\ bulb 2: R=\frac{120^2}{20}\\ R=720 ohms[/tex]
when miss connection occur we need to calculate the dissipated power from wash bulb
a. for bulb 1 with R=7.2ohms and V=120v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{120^2}{7.2}\\ P=2000W[/tex]
b. for bulb 2 with R=720 ohms and V=12v,
the power is calculated as
[tex]P=\frac{v^2}{R}\\ P=\frac{12^2}{720}\\ P=0.2W\\[/tex]
c. Bulb 1 is more likely to burn out because it is operating above the rated power.
The electron drift speed in a metal wire is exceedingly slow. Yet, when you turnon the light switch the light begins to illuminate almost immediately. Explain whyt his is not a paradox?
Explanation:
The misunderstanding here is that the thing that turns on the light bulb is not the same electrons near the light switch. So, the electrons near the switch is not moving all the way across the circuit instantly. The electrons are distributed across the wire. When the light switch is turned on, the circuit is connected and there is a potential difference between the bulb and the source. This potential difference creates an electric field, and free electrons move under the influence of this electric field according to Coulomb's Law. When they start to move the electrons closest to the bulb causes the bulb to glow.
So, the important factor here is not the drift velocity of the electrons but the number of electrons and the strength of the electric field.
Incandescent light bulbs contain a metallic filament inside. Metallic systems have allowed energy levels in a continuous range of energy, so electrons can make transitions of any energy within that range. In our lab, we will connect a light bulb to a variable AC voltage source (a Variac), which can deliver 0-140 V to the filament. The higher the voltage, the hotter we make the temperature, and the more energy we are giving the electrons in the metal, Suppose that at 20 V the filament is a dull red color, which means that most of the photons being emitted are red. When we increase the voltage, what color of light do you expect the filament to emit?
Answer:
Since at 20V most of the photons released are red then when the voltage keeps increasing the hotter the filament will be, therefore the color of light will be bright red.
Explanation:
The higher the energy the more the electrons in the molecules of the object will be excited, and when they de-excite to their ground states they release energy in the form of infrared light. The increase in voltage and higher temperatures make the object release brighter color and sometimes at the highest temperatures +1400 degrees Celsius, the color glows hot white.
The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtain an output voltage VO = 2.0 V. If a current of 1 mA is drawn away from the output terminal by a load, what is the change in output voltage?
Answer:
[tex] v = \frac{2 V}{3}= 0.667 v[/tex]
Since we have identical diodes we can use the equation:
[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]
And replacing we have:[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]
Since we know that 1 mA is drawn away from the output then the real value for I would be
[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]
And for this case the value for [tex] v_D[/tex] would be:
[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]
And the output votage on this case would be:
[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]
And the net change in the output voltage would be:
[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]
Explanation:
For this case we have the figure attached illustrating the problem
We know that the equation for the current in a diode id given by:
[tex] I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}[/tex]
For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode [tex] v_1 + v_2 + v_3= 2[/tex] and each voltage is the same v for each diode, so then:
[tex] v = \frac{2 V}{3}= 0.667 v[/tex]
Since we have identical diodes we can use the equation:
[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]
And replacing we have:
[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]
Since we know that 1 mA is drawn away from the output then the real value for I would be
[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]
And for this case the value for [tex] v_D[/tex] would be:
[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]
And the output votage on this case would be:
[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]
And the net change in the output voltage would be:
[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]
The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.6 cm apart with a 20 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.a)What is the electric field strength between the plates?
b)With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: the exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.
Answer:
(a) E = 1.25×10⁶ V/m
(b) v = 83.8×10⁶ m/s
Explanation:
Please see attachment below.
qV = electric potential energy of the electron. This is the energy converted into the kinetic energy of the motion of the electron through the plates. Kinetic energy is 1/2×mv². Where m = mass of electron = 9.11×10-³¹ kg.
q = 1.6×10-¹⁹ C.
So qV = 1/2mv² so rearranging and substituting the given quantities give the speed of the electrons.
Final answer:
The electric field strength between the plates is 1,250,000 N/C. Calculating the exit speed of an electron involves using the charge of the electron and the potential difference, but the accurate computation requires relativity due to the high velocity.
Explanation:
The student has asked two questions regarding the operation of an electron gun in an old TV picture tube: calculating the electric field strength between the plates and determining the exit speed of an electron.
Electric Field Strength Between the Plates
The electric field strength (E) between two parallel plates can be calculated using the formula E = V/d, where V is the potential difference between the plates, and d is the distance between them. With V = 20,000 volts (20 kV) and d = 0.016 meters (1.6 cm), E = 20,000 / 0.016 = 1,250,000 N/C.
Exit Speed of an Electron
To find the exit speed of an electron, we use the formula KE = eV, where KE is the kinetic energy of the electron, e is the charge of the electron (1.6 x 10⁻¹⁹ C), and V is the potential difference. The exit speed can then be derived from kinetic energy using the formula KE = 1/2 mv² leading to a theoretical calculation. However, the exit speed is significantly high, and thus, actually calculating it accurately involves considerations of relativity.
When a rattlesnake strikes, its head accelerates from rest to a speed of 28 m/s in 0.65 seconds. Assume for simplicity that the only moving part of the snake is its head of mass 170 g. How much (average) power does the rattlesnake need to accelerate its head that fast
Answer:
102.52 W
Explanation:
Power: This ca be defined as the rate at which energy is used up.
The S.I unit of power is Watt(W).
From the question,
P = E/t.............. Equation 1
Where P = average power of the snake head, E = energy of the snake, t = time.
But,
E = 1/2mv²............ Equation 2
Where m = mass of the head, v = velocity of the snake head.
Substitute equation 2 into equation 1
P = 1/2mv²/t ............ Equation 3
Given: m = 170 g = 0.17 kg, v = 28 m/s, t = 0.65 s.
Substitute into equation 3
P = 1/2(0.17)(28²)/0.65
P = 102.52 W.
Hence the average power of the snake = 102.52 W
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3×106N/C of the field causes a spark.
Answer:
[tex]9.56\cdot 10^{-7} C[/tex]
Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma=\frac{Q}{A}[/tex] is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
[tex]\epsilon_0 = 8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is
[tex]A=L^2=(0.19)^2=0.036m^2[/tex]
Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:
[tex]E=3\cdot 10^6 N/C[/tex]
Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
[tex]E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C[/tex]
A 1 kg object is initially 20m above the ground and rises to a height of 40m. Liz and Connor each measure its position, but each of them uses a different coordinate system. Select the choice that best represent the Δ P E as measured by the two scientists. Where we define Δ P E = P E f i n a l − P E i n i t i a l . Assume g = 10 m / s 2 .
The change in potential energy as the object rises from 20m to 40m is 200 Joules. This result is the same regardless of the coordinate system used.
Explanation:The change in potential energy, Δ P E, can be calculated using the formula Δ P E = P E f i n a l − P E i n i t i a l. Here, P E f i n a l is the final potential energy when the object is at 40m, and P E i n i t i a l is the initial potential energy when the object is at 20m. To get these values, we use the formula for gravitational potential energy: P E = m g h, where m is mass, g is the acceleration due to gravity, and h is the height above ground.
So, P E f i n a l = m g h = 1kg * 10 m/s² * 40m = 400 J (Joules), and P E i n i t i a l = m g h = 1kg * 10 m/s² * 20m = 200 J. Hence, Δ P E = P E f i n a l − P E i n i t i a l = 400 J - 200 J = 200 J. Regardless of their coordinate systems, both Liz and Connor should get the same Δ P E, assuming they've correctly identified the initial and final heights.
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The change in potential energy (ΔPE) as measured by both scientists would be 200J. This is calculated using the formula ΔPE = PEfinal - PEinitial, and substituting given values for mass, gravity, and height.
Explanation:In this scenario, the change in potential energy (ΔPE) is determined by the difference in heights and the weight of the object. We calculate ΔPE with the formula ΔPE = PEfinal − PEinitial. Given that PE is calculated as m*g*h where m is mass, g is gravity (which we assume as 10m/s² here), and h is height, we can substitute these values in. Initially, the object is at 20m, so PEinitial is m*g*h = 1kg*10m/s²*20m = 200 J (joules). Finally, it rises to 40m, so PEfinal = 1kg*10m/s²*40m = 400 J. Hence, ΔPE is PEfinal−PEinitial = 400J – 200J = 200J. Therefore, the change in potential energy, as measured by both Liz and Connor, would be 200J regardless of their coordinate systems.
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Explain why the particle in a box and the harmonic oscillator are useful models for quantum mechanical systems: what chemically significant systems can they be used to represent
Answer:
Because it's used to find definite solutions to More complex quantum mechanical system. Harmonic oscillator helps in determining motion of small mass if strings.
Explanation:
The particle box is very precise when using it to calculate or find definite solutions in complex quantum mechanics in which particles might be trapped in a regions of electric potentials. Since we're dealing with electric potentials here, the harmonic oscillator is needed because it has to do squarely wit potential energy which is given as
V(x) = 0.5kx²
Where k is force constant and x is distance.
Final answer:
The particle in a box and the harmonic oscillator models are useful for understanding quantum mechanical systems. They can be used to represent various chemically significant systems, including blackbody radiators, atomic and molecular spectra, and optoelectronic devices.
Explanation:
The particle in a box and the harmonic oscillator are useful models for quantum mechanical systems because they provide insights into the behavior of particles in confined spaces and under the influence of potential energy. The particle in a box model describes a particle free to move in a small space surrounded by impenetrable barriers, and can be used to represent systems such as blackbody radiators, atomic and molecular spectra, and optoelectronic devices. The harmonic oscillator model, on the other hand, represents systems with periodic motion, such as molecular vibrations and wave packets in quantum optics.
A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.40 m/s2. How far has the train traveled up the incline after 7.30 s
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
[tex]V_{x} = mt + V_{o}[/tex] where m is the slope
Comparing equation (1) and (2)
[tex]V = V_{x}[/tex]
a = m
[tex]U = V_{o}[/tex]
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
[tex]V_{x} = -1.40t + 22[/tex]
[tex]V_{x} = -1.40(7.30) + 22[/tex]
[tex]V_{x} = -10.22 + 22[/tex]
[tex]V_{x} = 11. 78 m/s[/tex]
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
[tex][\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)][/tex]
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
123.297 m
Explanation:
A train, traveling at a constant speed of 22.0 m/s,
v = 22.0 m/s
the train slows down with a constant acceleration of magnitude 1.40 m/s².
[tex]a_s[/tex] = -1.4 m/s²
How far has the train traveled up the incline after 7.30 s
t =7.30 s
We can calculate the distance traveled up the incline after 7.30 s by using the formula:
[tex]x_f =x_i+v_xt+\frac{1}{2}a_st^2[/tex]
where;
[tex]x_f[/tex] = the distance traveled up
[tex]x_i[/tex] = 0
[tex]v_x[/tex] = speed of the train
[tex]a_s[/tex] = deceleration
t = time
Substituting our data; we have:
[tex]x_f = 0+(22m/s)(7.30s)+\frac{1}{2}(-1.4m/s^2)(7.30s)[/tex]
[tex]x_f =16.06 -37.303[/tex]
[tex]x_f[/tex] = 123.297 m
An airplane propeller is 2.18 m in length (from tip to tip) and has a mass of 102 kg. When the airplane's engine is first started, it applies a constant net torque of 1950 N⋅m to the propeller, which starts from rest.What is the angular acceleration of the propeller? Treat the propeller as a slender rod.
Answer:
Explanation:
Moment of inertia of propeller
= (1/12) m l²
=( 1 / 12) x 102 x 2.18²
= 40.4 kgm²
torque = 1950 N-m
angular acceleration
= torque / moment of inertia
= 1950 / 40.4
= 48.26 rad / s²
Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 V. However, the battery delivers only 2.5 V. Therefore, we decide to charge two equal value capacitors in parallel from the 2.5V battery and then switch the capacitors in series with the heart during the 1ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout its 1ms duration
Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
Final answer:
To find the minimum capacitance for the cardiac pacemaker circuit to maintain the pulse amplitude between 4.9V and 5.0V over 1ms, use the discharging formula for a capacitor and solve for the capacitance with the given resistance and time.
Explanation:
The problem involves designing a cardiac pacemaker circuit where two capacitors charged to 2.5V in parallel are switched to series to deliver a pulse of 5V across a 500-ohm resistor, which models the heart. The question asks for the minimum value of the capacitance needed to ensure the voltage does not drop below 4.9V over the 1ms duration of the pulse.
To solve for the capacitance (C), we use the formula for the discharging of a capacitor through a resistor (Q = Q0e-(t/RC)), where Q is the charge at time t, Q0 is the initial charge, R is the resistance, and C is the capacitance. The minimum voltage (V min) we want after 1ms is 4.9V, which will be across two capacitors in series, effectively having a 5.0V drop across two capacitances when fully charged. The effective capacitance when two identical capacitors are in series is half that of one capacitor, so we need to calculate for C based on an effective voltage drop from 5.0V to 4.9V across a single capacitor, equivalent to the heart resistance of 500 ohms, over 1ms.
We rearrange the discharging formula to solve for C:
V = V 0 e -(t/RC)
Now we solve for C:
C = -t / (R*ln(V/V 0 )
Plugging in the values:
C = -0.001s / (500Ω*ln(4.9V/5.0V))
This calculation will give us the minimum capacitance value to ensure the pulse amplitude stays between 4.9V and 5.0V during the 1ms pulse duration.
3. A uniformly charged ring of radius 10.0 cm has a total T charge of 75.0 mC. Find the electric field on the axis of the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and (d) 100 cm from the center of the ring.
To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,
[tex]E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}[/tex]
Here,
[tex]k_e[/tex] = Coulomb constant
q = Charge
x = Distance to the axis
r = Distance between the charges
Our values are given as,
[tex]r = 10.0cm = 10*10^{-2} m[/tex]
[tex]q = 75\mu C = 75*10^{-6} C[/tex]
[tex]x_a = 1.00cm[/tex]
[tex]x_b = 5.00cm[/tex]
[tex]x_c = 30cm[/tex]
[tex]x_d = 100cm[/tex]
So applying this to our 4 distances, we have
PART A)
[tex]E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_a = 6.64*10^6N/C \hat{i}[/tex]
PART B)
[tex]E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_b = 24.1*10^6N/C \hat{i}[/tex]
PART C)
[tex]E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_c = 6.39*10^6N/C \hat{i}[/tex]
PART D)
[tex]E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]
[tex]E_d = 0.664*10^6N/C \hat{i}[/tex]
A circular copper wire is put in tension under a weight of 7000N. What is the ratio of its diameter after and before the load is applied, if the initial cross section of the wire is 0.01m2 and its Poisson ratio is 0.3
Answer:
[tex]\frac{d_f}{d} =0.9999983[/tex]
Explanation:
Given:
force applied on the copper wire, [tex]F=7000\ N[/tex]cross sectional area of the wire, [tex]a=0.01\ m^2[/tex]Poisson's ratio, [tex]\mu=0.3[/tex]we have, Young's modulus, [tex]E=128\times 10^3\ MPa[/tex]Stress induced due to the applied force:
[tex]\sigma=\frac{F}{a}[/tex]
[tex]\sigma=\frac{7000}{0.01}[/tex]
[tex]\sigma=700000\ Pa=0.7\ MPa[/tex]
Now the longitudinal strain:
[tex]\epsilon=\frac{\sigma}{E}[/tex]
[tex]\epsilon=\frac{0.7}{128\times 10^3}[/tex]
[tex]\epsilon=5.468\times 10^{-6}[/tex]
Now from the relation of Poisson's ratio:
[tex]\mu=\frac{\nu}{\epsilon}[/tex]
where:
[tex]\nu=[/tex] lateral strain
[tex]0.3=\frac{\nu}{5.468\times 10^{-6}}[/tex]
[tex]\nu=1.6406\times 10^{-6}[/tex] ..................(1)
Now we find the diameter of the wire:
[tex]a=\pi.\frac{d^2}{4}[/tex]
[tex]0.01=\pi\times \frac{d^2}{4}[/tex]
[tex]\frac{0.04}{\pi} =d^2[/tex]
[tex]d=0.1128\ m=112.8\ mm[/tex]
When the tensile load is applied its diameter decreases:
The lateral strain is also given as,
[tex]\nu=\frac{\Delta d}{d}[/tex]
[tex]1.6406\times 10^{-6}=\frac{\Delta d}{112.8}[/tex]
[tex]\Delta d=0.000185\ mm[/tex]
Now the final diameter will be:
[tex]d_f=d-\Delta d[/tex]
[tex]d_f=112.8-0.000185[/tex]
[tex]d_f=112.799815\ mm[/tex]
Now the ratio:
[tex]\frac{d_f}{d} =\frac{112.799815}{112.8}[/tex]
[tex]\frac{d_f}{d} =0.9999983[/tex]
A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t)=4.50cmcos[(19.5s−1)t−π/8].(a) What is the spring constant of the spring?
Answer:
465.6 N/m
Explanation:
We are given that
F=12 N
[tex]y(t)=4.50cos\left \{(19.5s^{-1}t-\frac{\pi}{8}\right \}[/tex]
We have to find the spring constant of the spring.
[tex]F=mg[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Using the formula
[tex]12=m\times 9.8[/tex]
[tex]m=\frac{12}{9.8}[/tex]kg
Compare the given equation with
[tex]y(t)=Acos(\omega t-\phi)[/tex]
We get [tex]\omega=19.5[/tex]
[tex]k=m\omega^2[/tex]
Using the formula
Spring constant,[tex]k=\frac{12}{9.8}\times (19.5)^2=465.6 N/m[/tex]
The spring constant (k) for the object executing simple harmonic motion is approximately 463.29 N/m, calculated using the provided angular frequency and the mass of the object.
Explanation:To find the spring constant (k) of the spring for an object executing simple harmonic motion (SHM), we use the equation of motion provided:
y(t) = 4.50 cm cos[(19.5 s-1)t - π/8]
From the equation of motion, we can see that the angular frequency (ω) is 19.5 s-1. In SHM, the angular frequency ω is related to the spring constant k and the mass m by the following equation:
ω = sqrt(k/m)
To calculate k, we rearrange the equation to solve for k:
k = mω2
First, we find the mass (m) by using the weight of the object (12.0 N):
m = weight / g = 12.0 N / 9.81 m/s2 ≈ 1.22 kg
Then we calculate the spring constant (k):
k = (1.22 kg)(19.5 s-1)2 ≈ 463.29 N/m
Therefore, the spring constant of the spring is approximately 463.29 N/m.
A girl whirls a stone in a horizontal circle 1.50 m above the ground by means of a string 165 cm long. The string breaks, and the stone flies of horizontally and strike the ground 6.90 m away. What is the centripetal acceleration of the stone while in circular motion
Answer:
95.5ms-2
Explanation:
First we obtain the time taken for the motion from the equation of motion. Secondly we obtain the horizontal velocity of the stone. This can now be used to calculate the centripetal acceleration. Note that the length of the chord is the radius of the circle around which the stone moves.
All details are contained in the detailed step-by-step solution attached to this answer.
A disk of radius R (Fig. P25.73) has a nonuniform surface charge density s 5 Cr, where C is a constant and r is measured from the center of the disk to a point on the surface of the disk. Find (by direct integration) the electric potential at P.
Answer:
Explanation:
dV = k (dq)/d
dV = k (sigma da)/d
dV = k (5 C r (r dr d(phi)))/d
d = sqrt(r^2 + x^2)
dV = 5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)
==> V = int{5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C int{(r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C (2 pi) int{(r^2/sqrt(r^2+x^2)) dr} ; from r=0 to r=R
==> V = 5 k C (2 pi) (1/2) (R sqrt(R^2+x^2) - x^2 ln(sqrt(r^2+x^2) + r) + x^2 Ln(x))
In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x10-27 kg). Assume that the radius of the orbit is 2.5 Angstroms (10-10 m). (a) What is the magnitude of the gravitational force of attraction between electron and proton? AN (b) What is the magnitude of the electric force of attraction between electron and proton? N
Answer:
(a) [tex]F_g=1.62*10^{-48}N[/tex]
(b) [tex]F_e=3.68*10^{-9}N[/tex]
Explanation:
(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:
[tex]F_g=-G\frac{m_1m_2}{r^2}[/tex]
Where G is the Cavendish gravitational constant, [tex]m_1[/tex] and [tex]m_2[/tex] are the masses of the electron and the proton respectively and r is the distance between them:
[tex]F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N[/tex]
The minus sing indicates that the force is repulsive. Thus, its magnitude is:
[tex]F_g=1.62*10^{-48}N[/tex]
(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:
[tex]F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N[/tex]
Its magnitude is:
[tex]F_e=3.68*10^{-9}N[/tex]
Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an angle of 45 degrees. The ball is 2.0 m off the ground when it leaves his hand, and hits the backboard 3.5 m off the ground. Neglecting air friction, about how long is the ball in flight?
Answer:
The time of flight of the ball is 1.06 seconds.
Explanation:
Given [tex]\Delta x=7\ m[/tex]
[tex]\theta=45 \°[/tex]
Also, [tex]\Delta y=(3.5-2)=1.5\ m[/tex]
[tex]a_x=0\ and\ a_y=-9.81\ m/s^2[/tex]
Let us say the velocity in the x-direction is [tex]v_x[/tex] and in the y-direction is [tex]v_y[/tex]. And acceleration in the x-direction is [tex]a_x[/tex] and in the y-direction is [tex]a_y[/tex].
Also, [tex]\Delta x\ and\ \Delta y[/tex] is distance covered in x and y direction respectively. And [tex]t[/tex] is the time taken by the ball to hit the backboard.
We can write [tex]v_x=v_0cos(45)\ and\ v_y=v_0sin(45)[/tex]. Where [tex]v_0[/tex] is velocity of ball.
Now,
[tex]\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt[/tex]
[tex]\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}[/tex]
Also,
[tex]\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s[/tex].
Plugging this value in
[tex]t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}[/tex]
[tex]t=1.06\ seconds[/tex]
So, the time of flight of the ball is 1.06 seconds.
Two light bulbs are operated at a potential difference of 110 V. Light bulb A produces 60 W of power and light bulb B produces 100 W of power. Which statement below is correct?
a. The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.
b. The 60 W bulb has a smaller resistance and smaller current than the 100 W bulb.
c. The 60 W bulb has a greater resistance and greater current than the 100 W bulb.
d. The 60 W bulb has a smaller resistance and greater current than the 100 W bulb.
Answer:a
Explanation:
Given
Potential difference [tex]V=110\ V[/tex]
Power of bulb A [tex]P_A=60\ W[/tex]
Power of bulb B [tex]P_B=100\ W[/tex]
If voltage is same for both the bulbs then Power is given by
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]P_A=\dfrac{(110)^2}{R_A}[/tex]
[tex]60=\dfrac{110^2}{R_A}[/tex]
[tex]R_A=201.66\ \Omega[/tex]
similarly
[tex]R_B=\dfrac{110^2}{100}[/tex]
[tex]R_B=121\ \Omega[/tex]
[tex]R_A>R_B[/tex]
so current in bulb A is smaller than B
Thus the 60 W bulb has a greater resistance.
Thus option (a) is correct
The correct statement is that a) the 60 W bulb has a greater resistance and smaller current than the 100 W bulb, since both bulbs are operated at the same voltage but the 60 W bulb has lower power.
To determine the correct statement about the resistance and current for two light bulbs operated at a potential difference of 110 V, we need to consider the power ratings of the bulbs and know that power (P) is the product of current (I) and voltage (V), P = IV. Furthermore, by using Ohm's law (V = IR), we can infer that Power can also be expressed as P = I2R or P = V²/R, which relates power to resistance (R).
Both bulbs operate at the same voltage (110 V), which means their power differences are due to differences in current and resistance.
For bulb A (60 W): P = V²/R, so R = V²/P = (110 V)2/60 W.
For bulb B (100 W): R = V²/P = (110 V)²/100 W.
Comparing these two, bulb A has higher resistance because it has lower power for the same voltage. Also, since P = IV, and bulb A has a lower power at the same voltage, it must also have a smaller current. Therefore, the correct answer is: a. The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 5.6 cm and rotates in a 0.24-T field. Coil 2 rotates in a 0.44-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2?
Answer:
4.14 cm.
Explanation:
Given,
For Coil 1
radius of coil, r₁ = 5.6 cm
Magnetic field, B₁ = 0.24 T
For Coil 2
radius of coil, r₂ = ?
Magnetic field, B₂ = 0.44 T
Using formula of maximum torque
[tex]\tau_{max}= NIAB[/tex]
Since both the coil experience same maximum torques
now,
[tex] NIA_1B_1 = NIA_2B_2[/tex]
[tex]A_1B_1 = A_2B_2[/tex]
[tex] r_1^2 B_1 = r_2^2 B_2[/tex]
[tex] 5.6^2\times 0.24= r_2^2\times 0.44[/tex]
[tex]r_2 = 4.14\ cm[/tex]
Radius of the coil 2 is equal to 4.14 cm.
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away
Answer:
0.82 mm
Explanation:
The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:
[tex]y_n=\frac{n \lambda D}{d}[/tex]
so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:
[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]
where;
n = 2
[tex]\lambda_1[/tex] = 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y_2[/tex] = 0.00276 m
[tex]y_2[/tex] = 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:
[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]
[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
[tex]\delta y = y_2-y^{'}_2[/tex]
[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)
[tex]\delta y[/tex] = 10 ⁻³ (0.82)
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m
[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]
[tex]\delta y[/tex] = 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Two infinite plane sheets with uniform surface charge densities are placed parallel to each other with separation d. in the region between the sheets, where does the total electric field have the greatest magnitude ? please explain
The density given implies that the total electric field will have a uniform magnitude.
What is density?Density simply means the degree of compactness that can be found in a substance.
It should be noted that for an infinite sheet of charge, the electric field will be perpendicular to the surface.
In this case, it doesn't depend on the distance of points and is uniform all through. Therefore, the total electric field will have a uniform magnitude.
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A system ______.
(A) is a fixed amount of matter and a control volume
(B) is a volume in space.
(C) is a volume in space and a control volume is a fixed amount of matter.
(D) and a control volume are both a fixed amount of matter.
(E) and a control volume are both a volume in space.
Answer:
(B) is a volume in space.
Explanation:
A system is the space chosen form the universe for the study and analysis.
There are 3 kinds of systems :
1) Open system:
This system can interact to surroundings in terms of both mass and energy transfer through its boundaries.
2) Closed system:
This type of system can only interact to the surrounding in terms of energy transfer through its boundary.
3) Isolated system:
This type of system cannot interact with the surrounding by any means and its energy remains constant with time under the observation.
It has insulated boundary.
A system in physics often represents a fixed set of physical interactions, whereas a control volume is a designated region in space chosen to study the balance of mass, energy, and momentum. Option (A) best defines these concepts.
Explanation:In physics, a system typically refers to a set of physical interactions that are being studied. This could include anything from a single particle to a complex machine. A control volume, on the other hand, is a specific volume in space. The essence of its definition is that it is a region in space chosen for study of the balance of mass, energy, and momentum. The boundaries of the control volume can be real or imagined. The control volume can also move and deform with time.
To answer the question, option (A) would be most accurate. A system is a fixed amount of matter, meaning it represents a particular set of physical elements. A control volume is a region in space which can encompass these elements.
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A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x10^7 m/s and experiences an acceleration of 2.00 x10^13 m/s^2. in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field for which the magnitude of the field is a minimium
Answer:
So, Magnitude of the field (B) = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
Explanation:
First of all, since acceleration is in positive x- direction, the magnetic field must be in negative y- direction.
We know that The magnitude of the Lorentz force F is; F = qvB sinθ
So, B = F/(qvsinθ)
F = ma.
Speed(v) = 1.00 x10^(7) m/s
acceleration (a) = 2.00 x10^(13) m/s^(2)
Mass of proton = 1.673 × 10^(-27) kilograms
q(elementary charge of proton) = 1.602×10^(−19)
Since right hand thumb rule, θ= 90°
So;B = [1.673 × 10^(-27) x 2.00 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]
So,B = 2.09 x 10^(-2) T.
It's in the negative direction since acceleration is in positive direction.
The magnitude of the magnetic field that the proton experiences is 2.08 x 10^-4 Tesla. The direction of the magnetic field, determined by the right-hand rule, is in the negative y-direction.
Explanation:The force experienced by a charged particle moving within a magnetic field, like the proton in your question, can be calculated using the Lorentz force law: F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Because the proton is moving perpendicular to the field, θ = 90°, and sinθ = 1. Therefore, F = qvB. To find the magnetic field, B, we can rearrange this formula to give B = F/qv.
Now, we know from Newton's second law that F = ma, so we can replace F in our formula with the proton's mass (m, which is 1.67 x 10^-27 kg for a proton) times the given acceleration (a), yielding B = ma/qv.
Substitute the given values for acceleration, proton's charge, and velocity into our equation, we get: B = (1.67 x 10^-27 kg)(2 x 10^13 m/s^2) / (1.60 x 10^-19 C)(1 x 10^7 m/s) = 2.08 x 10^-4 Tesla (T).
The direction of the magnetic field is given by the right-hand rule as discussed in Essential Knowledge 2.D.1. If you arrange your right hand such that your thumb points in the direction of the proton's velocity (positive z-direction) and your fingers curl in the direction of the force (positive x-direction), your palm points in the direction of the magnetic field, which would be in the negative y-direction.
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Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the same direction. Is there any point between the two wires where the magnetic field is zero?
Answer:
x= 0.077 m from the wire carrying 5.0 A current.
Explanation:
If the wire can be approximated as an infinite one, and we can neglect the diameter of the wire, we can find the magnetic field B at a distance d from the wire, with the following expression:[tex]B =\frac{\mu_{0} * I}{2*\pi*d}[/tex]
Due to the currents are in the same direction, this means that the magnetic field lines (taking the shape of circumferences) will have opposite directions between the wires.So, if we assume that at some distance from both wires, the magnetic field will be 0, we can write the following equation:[tex]\frac{\mu_{0} * I_{1}}{2*\pi*x} - \frac{\mu_{0} * I_{2} }{2*\pi*(d-x)} = 0[/tex]
where I₁ = 5.0A, I₂= 8.0A and d = 0.2 mSimplifying common terms, we can solve for x, as follows:[tex]\frac{I_{1} }{x} = \frac{I_{2} }{(d-x)} \\ \frac{5.0A}{x(m)} = \frac{8.0A}{(0.2m-x(m))}[/tex]
⇒ [tex]x =\frac{1m}{13} = 0.077 m = 7.7 cm[/tex]
A charge of 8.473 nC is uniformly distributed along the x-axis from −1 m to 1 m . What is the electric potential (relative to zero at infinity) of the point at 5 m on the x-axis? The value of the Coulomb constant is 8.98755 × 10^9 N · m2 /C^2 . Answer in units of V.
The electric potential at 5 meters on the x-axis due to a uniformly distributed charge of 8.473 nC along the x-axis from -1m to 1m is approximately 51.84 volts.
Explanation:To determine the electric potential at a point on the x-axis created by a uniformly distributed charge, we can use the formula for electric potential due to continuous charge distribution. For a continuous charge distribution along the x-axis, the electric potential (V) at a point (P) is given by the integral of [tex]\(k \cdot \frac{dq}{r}\), where \(k\)[/tex] is Coulomb's constant, (dq) is the small charge element, and \(r) is the distance between the charge element and the point (P).
In this scenario, we integrate the expression across the entire distribution of charge from -1m to 1m along the x-axis to find the total potential at point 5m on the x-axis. Since the charge is uniformly distributed, we can represent (dq) as [tex]\(\lambda dx\),[/tex]where [tex]\(\lambda\)[/tex] is the charge per unit length and \(dx\) is the small element of length along the axis.
By integrating this expression and plugging in the given values into the formula, we obtain the result of approximately 51.84 volts for the electric potential at the point 5m away from the charge distribution on the x-axis.
This value signifies the work done per unit charge to bring a positive test charge from infinity to the specified point, considering the influence of the continuous charge distribution along the x-axis.
"".
A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 5.10-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.If the box is initially at rest at x=0, what is its speed after it has traveled 14.0 m ?
Answer:
[tex]v=7.62\ m.s^{-1}[/tex]
Explanation:
Given:
initial position of the box, [tex]x=0\ m[/tex]final position of the box, [tex]x'=14\ m[/tex]mass of the box under the force, [tex]m=5.1\ m[/tex]initial speed of the box, [tex]u=0\ m.s^{-1}[/tex]function of force, [tex]F(x)=18-0.53x\ [N][/tex]where:
[tex]x=[/tex] distance in the +ve x-direction
We know:
[tex]F=m.a\\\Rightarrow a=\frac{F}{m}[/tex]
Now force change in force on the body:
[tex]F(x)=18-0.53(x'-x)[/tex]
[tex]F=18-0.53\times (14-0)[/tex]
[tex]F=10.58\ N[/tex]
Now the acceleration due to the force:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{10.58}{5.1}[/tex]
[tex]a=2.0745\ m.s^{-2}[/tex]
Now using equation of motion:
[tex]v^2=u^2+2a.x'[/tex]
[tex]v^2=0^2+2\times 2.0745\times 14[/tex]
[tex]v=7.62\ m.s^{-1}[/tex]