Answer:
1.42×10⁻¹⁶ N.
Explanation:
The force on a charge moving in a magnetic field is given as
F = qBvsin∅..................... Equation 1
Where F = Force on the charge, q = charge, B = Magnetic Field, v = speed, ∅ = angle between the magnetic field and the speed
Given: B = 0.78 T, v = 1140 m/s, ∅ = 90° ( Perpendicular) q = 1.60 × 10⁻¹⁹ C.
Substitute into equation 1
F = 0.78(1140)(1.60 × 10⁻¹⁹)sin90°
F = 1.42×10⁻¹⁶ N.
Hence the force on the charge = 1.42×10⁻¹⁶ N.
In a typical Van de Graaff linear accelerator, protons are accelerated through a potential difference of 20 MV. What is their kinetic energy if they started from rest? Give your answer in (a) eV, (b) keV,(c) MeV, (d) GeV, and (e) joules.
Answer:
a) 2 x10^7 eV
b) 2 x10^4 keV
c) 20 MeV
d) 0.02 Gev
e) 3.2 x 10^-12J
Explanation:
The potential difference = 20 x 10^6 V
The charge on the proton = 1.6 x10^-19
The work done to move the proton will be basically the proton will acquire if it accelerates.
Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19
=3.2 x 10^-12J or 2 x10^7 eV
2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev
Explanation:
Below is an attachment containing the solution.
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
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.
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.
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
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.
Learn more about Magnetic Field here:https://brainly.com/question/36936308
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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
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]
At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.
Answer:
(a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Explanation:
Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.
Given that,
Mass of steel block = 2.30 kg
Mass of ball = 0.500 kg
Length of cord = 50.0 cm
We need to calculate the initial speed of the ball
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=mgl[/tex]
[tex]v=\sqrt{2gl}[/tex]
Put the value into the formula
[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]
[tex]u=3.13\ m/s[/tex]
The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]
The initial speed of the block [tex]u_{2}=0[/tex]
(a). We need to calculate the speed of the ball after collision
Using formula of collision
[tex]v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]
[tex]v_{1}=-2.01\ m/s[/tex]
Negative sign shows the opposite direction of initial direction.
(b). We need to calculate the speed of the block after collision
Using formula of collision
[tex]v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}[/tex]
Put the value into the formula
[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]
[tex]v_{2}=1.11\ m/s[/tex]
Hence, (a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
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.
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.
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
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.
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))
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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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]
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.
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Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height?
a. Object 1 rises to the greater height because it weighs less.
b. Object 2 rises to the greater height because it possesses a larger amount of kinetic energy.
c. Object 2 rises to the greater height because it contains more mass.
d. Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy.
e. The two objects rise to the same height.
Answer:
Option E is correct.
The two objects rise to the same height.
Explanation:
Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.
Let the mass, velocity and height the object will rise to be m, v and h respectively.
Note that V = v
Using the work energy theorem
The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)
(The final kinetic energy = 0 J because the object comes to rest at the final point)
Work done on the bigger object = work done by all the forces acting on the body
But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH
(-MV²/2) = - MgH
H = (V²/2g)
For the small body,
Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)
Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh
(-mv²/2) = - mgh
h = (v²/2g)
Since V = v as given in the question (Both bodies have the same speeds)
H = h = (V²/2g) = (v²/2g)
Hope this Helps!!!
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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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.
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]
To model a tornado, superimpose the stream function for a vortex (Eq. 6.91) and a source (Eq. 6.83). Assume that the circulation is 8000 m2 /s and the pressure at the radius of 40 m is 2 kPa less than atmospheric
Answer:
The stream function for this potential flow is -4696.8
Explanation:
Given that,
Circulation = 8000 m²/s
Radius = 40 m
Pressure = 2 kPa
Suppose the determine the stream function for this potential flow
We need to calculate the stream function
Using formula of stream function
[tex]\Psi=-(\dfrac{\Gamma}{2\pi})ln(r)[/tex]
Where, Γ = circulation
r = radius
Put the value into the formula
[tex]\Psi=-\dfrac{8000}{2\pi}\times ln(40)[/tex]
[tex]\Psi=-4696.8[/tex]
Hence, The stream function for this potential flow is -4696.8
Old cannons were built on wheeled carts, both to facilitate moving the cannon and to allow the cannon to recoil when fired. When a 150 kg cannon and cart recoils at 1.5 m/s, at what velocity would a 10 kg cannonball leave the cannon?
Answer:
22.5 m/s
Explanation:
Applying Newton's third law of motion
Momentum of the cannon and cart = momentum of the cannonball
MV = mv..................... Equation 1
Where M = mass of the cannon and the cart, V = Recoil velocity of the cannon and the cart, m = mass of the cannonball, v = velocity of the cannonball
make v the subject of the equation
v = MV/m............. Equation 2
Given: M = 150 kg, V = 1.5 m/s, m = 10 kg
Substitute into equation 2
v = 150(1.5)/10
v = 22.5 m/s
Hence the cannonball leave the cannon with a velocity of 22.5 m/s
Answer:
v2 = 22.5 m/s
Explanation:
Momentum is how hard to stop or turn a moving object . Generally, momentum measures mass in motion. Momentum is a vector quantity. Mathematically,
p = mass × velocity
The total momentum of an isolated system of bodies remains constant.
mometum before = 0
mass of the canon (m1) = 150 kg
mass of the ball (m2) = 10 kg
velocity of the ball (v2) = ?
velocity of the cannon(v1) = 1.5 m/s
momentum after = momentum before
m2v2 + m1v1 = 0
10v2 = 150 × 1.5
10v2 = 225
divide both sides by 10
v2 = 225/10
v2 = 22.5 m/s
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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Consider a sphere, an infinitely long cylinder, and a plane of infinite length and width (a, b and c below). Imagine that you can hover above each one in your own personal helicopter. In which case do you have the most freedom to move about without your view of the object changing? In other words, for each case consider if there are directions that you can move in without the objects distance or orientation, relative to you, changing.
Answer:
A plane of infinite length and width.
Explanation:
For the case of the sphere, we should consider spherical coordinates.
You can move around the sphere without changing your distance from the center of the sphere, however you can alter your azimuthal angle and polar angle, since they are symmetric in spherical coordinate system.
r --> Cannot change
Φ --> Free to change
θ --> Free to change
For the case of the infinite cylinder, we should consider cylindrical coordinates.
You can change your height and angular coordinate, but you cannot change your distance from the axis of the cylinder.
r --> Cannot change
θ --> Free to change
z --> Free to change
For the case of the infinite plane, we should consider cartesian coordinates.
Since the length and width of the plane is infinite, we cannot recognize whether we are getting closer or further away.
x --> Free to move
y --> Free to move
z --> Free to move
Therefore, in the case of infinite plane you have the most freedom to move about without your view of the object changing.
Final answer:
An infinite plane allows the most freedom of movement without changing the view, due to its infinite symmetry. In contrast, a sphere and a cylinder have more limited symmetrical properties, resulting in a restricted freedom of movement to keep the view unchanged.
Explanation:
When comparing the freedom to move above a sphere, an infinitely long cylinder, and an infinite plane, the plane provides the most freedom without changing your view of the object. On a sphere, any move results in a different orientation or distance from the surface. With the cylinder, movement along the axis does not change the view, but any other direction does. However, the infinite plane allows movement in any direction on a two-dimensional plane without changing the viewed geometry or orientation.
From a mathematical perspective, this question revolves around symmetry and geometric invariance under transformation. The infinite plane exhibits infinite symmetry; therefore, no matter how you move parallel to its surface, the view remains unchanged. Conversely, a sphere has rotational symmetry around its center, and a cylinder along its axis, limiting the directions in which one can move without altering the perceived distance or orientation.
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.
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]
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.
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 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.