Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.
Charge and Charge DensityA certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.
Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.
[tex]\sigma = \dfrac {Q}{A}[/tex]
[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]
[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]
Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].
Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.
[tex]Q' = \sigma A'[/tex]
Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.
For the closed Gaussian Surface, Flux is given below.
[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]
Hence the charge can be written as,
[tex]Q' = \phi\epsilon_\circ[/tex]
So the charge can be given as below.
[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]
Then the surface area of the Gaussian surface is given below.
[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]
Substituting the values in the above equation,
[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]
[tex]A' =0.006076\;\rm m^2[/tex]
[tex]A' = 60.76 \;\rm cm^2[/tex]
Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.
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what is drift speed ? {electricity}
Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
Explanation:
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex + (1/ 2)ez ? Are there any shear stresses acting on this surface?
Complete Question:
Given [tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex] at a point. What is the force per unit area at this point acting normal to the surface with[tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex] ? Are there any shear stresses acting on this surface?
Answer:
Force per unit area, [tex]\sigma_n = 28 MPa[/tex]
There are shear stresses acting on the surface since [tex]\tau \neq 0[/tex]
Explanation:
[tex]\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right][/tex]
equation of the normal, [tex]\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z[/tex]
[tex]\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]
Traction vector on n, [tex]T_n = \sigma \b n[/tex]
[tex]T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right][/tex]
[tex]T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right][/tex]
[tex]T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z[/tex]
To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.
[tex]\sigma_n = T_n . \b n[/tex]
[tex]\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa[/tex]
If the shear stress, [tex]\tau[/tex], is calculated and it is not equal to zero, this means there are shear stresses.
[tex]\tau = T_n - \sigma_n \b n[/tex]
[tex]\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z[/tex]
[tex]\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa[/tex]
Since [tex]\tau \neq 0[/tex], there are shear stresses acting on the surface.
Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other
Answer:
magnitude of the resultant of forces is 11.45 N
Explanation:
given data
force F1 = 6N
force F2 = 8N
angle = 240°
solution
we get here resultant force that is express as
F(r) = [tex]\sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}[/tex] ..............1
put here value and we get
F(r) = [tex]\sqrt{6^2+8^2+2\times 6\times 8 \times cos240}[/tex]
F(r) = 11.45 N
so magnitude of the resultant of forces is 11.45 N
A ball is dropped from the top of an eleven-story building to a balcony on the ninth floor. In which case is the change in the potential energy associated with the motion of the ball the greatest
Answer:
at the top of the 9 story building i think
Explanation:
When the ball starts to move, its kinetic energy increases and potential energy decreases. Thus the ball will experience its maximum potential energy at the top height before falling.
What is potential energy?Potential energy of a massive body is the energy formed by virtue of its position and displacement. Potential energy is related to the mass, height and gravity as P = Mgh.
Where, g is gravity m is mass of the body and h is the height from the surface. Potential energy is directly proportional to mass, gravity and height.
Thus, as the height from the surface increases, the body acquires its maximum potential energy. When the body starts moving its kinetic energy progresses and reaches to zero potential energy.
Therefore, at the sate where the ball is at the top of the building it have maximum potential energy and then changes to zero.
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What if a solid cylinder of mass M = 2.50 kg, radius R = 2.18 cm, and length L = 2.7 cm, is rolling down from rest instead? With h = 79.60 m and x = 4.64 m, what is the center of mass velocity when the cylinder reaches the bottom?
Answer:
The center of mass velocity is [tex]v = 32.25 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the cylinder is [tex]m = 2.50 \ kg[/tex]
The radius is [tex]r = 2.18 \ cm = 0.0218 \ m[/tex]
The length is [tex]l = 2.7 \ cm = 0.027 \ m[/tex]
The height of the plane is h = 79.60 m
and the distance covered is [tex]d = 4.64 \ m[/tex]
The center of mass velocity o the cylinder when it reaches the bottom is mathematically represented as
[tex]v = \sqrt{\frac{4gh}{3} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{4 * 9.8 * 79.60}{3} }[/tex]
[tex]v = 32.25 \ m/s[/tex]
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c
Answer:
0.85c
Explanation:
Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²
Rest mass of proton [tex]M_{0P}[/tex] = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV
for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV
Recall that the rest energy, and the total energy are related by..
[tex]E[/tex] = γ[tex]E_{0}[/tex]
which can be written in this case as
[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1
where [tex]E[/tex] = total energy of the kaon, and
[tex]E_{0}[/tex] = rest energy of the kaon
γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]
where [tex]\beta = \frac{v}{c}[/tex]
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2
where [tex]E_{K}[/tex] is the total energy of the kaon, and
[tex]E_{0P}[/tex] is the rest energy of the proton.
From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89
1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1
squaring both sides, we get
3.57( 1 - [tex]\beta^{2}[/tex]) = 1
3.57 - 3.57[tex]\beta^{2}[/tex] = 1
2.57 = 3.57[tex]\beta^{2}[/tex]
[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72
[tex]\beta = \sqrt{0.72}[/tex] = 0.85
but, [tex]\beta = \frac{v}{c}[/tex]
v/c = 0.85
v = 0.85c
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the surface by the block
Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)
Answer:
θ₁ = 85.5º θ₂ = 12.98º
Explanation:
Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.
Let's write the equations for motion for this point
X axis
x = v₀ₓ t
x = v₀ cos θ t
Y axis
y = [tex]v_{oy}[/tex] t - ½ g t2
y = v_{o} sin θ t - ½ g t²
let's substitute the values
100 = 80 cos θ t
15 = 80 sin θ t - ½ 9.8 t²
we have two equations with two unknowns, so the system can be solved
let's clear the time in the first equation
t = 100/80 cos θ
15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²
15 = 100 tan θ - 7.656 sec² θ
we can use the trigonometric relationship
sec² θ = 1- tan² θ
we substitute
15 = 100 tan θ - 7,656 (1- tan² θ)
15 = 100 tan θ - 7,656 + 7,656 tan² θ
7,656 tan² θ + 100 tan θ -22,656=0
let's change variables
tan θ = u
u² + 13.06 u + 2,959 = 0
let's solve the quadratic equation
u = [-13.06 ±√(13.06² - 4 2,959)] / 2
u = [13.06 ± 12.599] / 2
u₁ = 12.8295
u₂ = 0.2305
now we can find the angles
u = tan θ
θ = tan⁻¹ u
θ₁ = 85.5º
θ₂ = 12.98º
A long, straight wire with a circular cross section of radius R carries a current I. Assume that the current density is not constant across the cross section of the wire, but rather varies as J=αrJ=αr, where αα is a constant.
(a) By the requirement that J integrated over the cross section of the wire gives the total current I, calculate the constant αα in terms of I and R.
(b) Use Ampere’s law to calculate the magnetic field B(r) for (i) r≤Rr≤R and (ii) r≥Rr≥R. Express your answers in terms of I.
Answer: (a) α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]
(b) For r≤R: B(r) = μ_0.[tex](\frac{I.r^{2}}{2.\pi.R^{3}})[/tex]
For r≥R: B(r) = μ_0.[tex](\frac{I}{2.\pi.r})[/tex]
Explanation:
(a) The current I enclosed in a straight wire with current density not constant is calculated by:
[tex]I_{c} = \int {J} \, dA[/tex]
where:
dA is the cross section.
In this case, a circular cross section of radius R, so it translates as:
[tex]I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr[/tex]
[tex]I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}[/tex]
[tex]\alpha = \frac{3I}{2.\pi.R^{3}}[/tex]
For these circunstances, α = [tex]\frac{3I}{2.\pi.R^{3}}[/tex]
(b) Ampere's Law to calculate magnetic field B is given by:
[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]
(i) First, first find [tex]I_{c}[/tex] for r ≤ R:
[tex]I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr[/tex]
[tex]I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr[/tex]
[tex]I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}[/tex]
[tex]I_{c} = \frac{I.r^{3}}{R^{3}}[/tex]
Calculating B(r), using Ampere's Law:
[tex]\int\ {B} \, dl =[/tex] μ_0.[tex]I_{c}[/tex]
[tex]B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )[/tex].μ_0
B(r) = [tex](\frac{Ir^{3}}{R^{3}2.\pi.r})[/tex].μ_0
B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0
For r ≤ R, magnetic field is B(r) = [tex](\frac{Ir^{2}}{2.\pi.R^{3}} )[/tex].μ_0
(ii) For r ≥ R:
[tex]I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr[/tex]
So, as calculated before:
[tex]I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}[/tex]
[tex]I_{c} =[/tex] I
Using Ampere:
B.2.π.r = μ_0.I
B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0
For r ≥ R, magnetic field is; B(r) = [tex](\frac{I}{2.\pi.r} )[/tex].μ_0.
A number, which represents a property, amount, or relationship that does not change under certain situations is constant and further calculations as follows:
constant calculation:The Radius of the cross-section of the wire R
Current passing through the wire I
Current Density [tex]J = \alpha r[/tex]
Constant [tex]\alpha[/tex]
Distance of the point from the center [tex]r[/tex]
For part a)
Consider a circular strip between two concentric circles of radii r and r+dr.
Current passing through the strip [tex]dI =\overrightarrow J \times \overrightarrow{dA}[/tex]
[tex]\to\alpha r (2\pi r dr) cos 0^{\circ}[/tex]
Integration
[tex]\to I =2\pi \alpha \int^R_0 r^2\ dr =2\pi \alpha [r^3]^R_0=2\pi \alpha \frac{r^3}{3}\\\\\to \alpha = \frac{3I}{2\pi R^3}\\\\[/tex]
For part b)
The magnetic field at a point distance [tex]'r'^{(r \ \pounds \ R)}[/tex] from the center is B.
We have the value of the line integral of the magnetic field over a circle of radius ‘r’ given as
[tex]\oint \overrightarrow B \times \overrightarrow{dl} = \mu_0 I\\\\[/tex]
where ‘I’ is the threading current through the circle of radius ‘r’
[tex]\oint B \ dl \cos 0^{\circ} = \mu_0 [2\pi \alpha \frac{r^3}{3}]\\\\ B \int dl = \mu_0 [2\pi \frac{3I}{2\pi R^3} \frac{r^3}{3}]\\\\ B \cdot 2\pi r = \mu_0 I [\frac{r}{R}]^3\\\\ B = \frac{\mu_0}{2\pi} I [\frac{Ir^2}{R^3}]\\\\[/tex]
(ii) Similarly, we can calculate the magnetic field at the point at A distance ‘r’ where
[tex]\to r^3 R\\\\\to \int \overrightarrow{B} \overrightarrow{dl} = \mu_0\ I[/tex] [The threading current is the same]
[tex]\to \beta - 2\pi r = \mu_0 I[/tex] As (I)
[tex]\to \beta =\frac{\mu_o \ I}{2\pi \ r}[/tex]
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A crane lifts a 425 kg steel beam vertically a distance of 64 m. How much work does the crane do on the beam if the beam accelerates upward at 1.8 m/s2
Answer:
work done= 48.96 kJExplanation:
Given data
mass of load m= 425 kg
height/distance h=64 m
acceleration a= 1.8 m/s^2
The work done can be calculated using the expression
work done= force* distance
but force= mass *acceleration
hence work done= 425*1.8*64= 48,960 J
work done= 48.96 kJ
An electron (mass=9.11 X 10^-31kg) leaves a TV picture tube with zero initial speed and reaches the screen 1.90cm away at 3.00 X 10^6 m/s. Ignore gravity and find the net force. (1): 2.28 X 10^-19N (2): 2.28 X 10^-16N (3): None of the above
Answer:
2.16 × 10^-16N
Explanation:
The computation of the net force is shown below:
Data provided in the question
Electron mass = 9.11 × 10^-31kg
V_o = 0
V_f = 3.00 × 10^6 m/s
s = 1.90 cm i.e 1.9 × 10^-2
Based on the above information, the force is
As we know that
[tex]Force\ f = ma = \frac{mv^2}{2s}\\\\ = \frac{(9.11\times 10^{-31})(3\times 10^{6})^2}{2(1.9\times 10^{-2})}[/tex]
= 2.16 × 10^-16N
Hence, the last option is correct
Basically we applied the above formula to determine the net force
A man stands on a merry-go-round that is rotating at 2.5 rad/s. If the coefficient of static friction between the man’s shoes and the merry-go-round is µs = 0.5, how far from the axis of rotation can he stand without sliding?
Answer:
0.8 m
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing towards the center.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces in the centripetal direction:
∑F = ma
Nμ = m v²/r
Substitute and simplify:
mgμ = m v²/r
gμ = v²/r
Write v in terms of ω and solve for r:
gμ = ω²r
r = gμ/ω²
Plug in values:
r = (10 m/s²) (0.5) / (2.5 rad/s)²
r = 0.8 m
The distance (radius) from the axis of rotation which the man can stand without sliding is 0.784 meters.
Given the following data:
Angular speed = 2.5 rad/s.Coefficient of static friction = 0.5To determine how far (radius) from the axis of rotation can the man stand without sliding:
We would apply Newton's Second Law of Motion, to express the centripetal and force of static friction acting on the man.
[tex]\sum F = \frac{mv^2}{r} - uF_n\\\\\frac{mv^2}{r} = uF_n[/tex]....equation 1.
But, Normal force, [tex]F_n = mg[/tex]
Substituting the normal force into eqn. 1, we have:
[tex]\frac{mv^2}{r} = umg\\\\\frac{v^2}{r} = ug[/tex]....equation 2.
Also, Linear speed, [tex]v = r\omega[/tex]
Substituting Linear speed into eqn. 2, we have:
[tex]\frac{(r\omega )^2}{r} = ug\\\\r\omega ^2 = ug\\\\r = \frac{ug}{\omega ^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]r = \frac{0.5 \times 9.8}{2.5^2} \\\\r = \frac{4.9}{6.25}[/tex]
Radius, r = 0.784 meters
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Why can a magnetic monopole not exist, assuming Maxwell's Equations are currently correct and complete?
Answer:
Because closed magnetic field loops have to be formed between both ends of the magnet, a magnet will always have two poles.
Explanation:
Magnetic Monopoles do not exist in nature because a magnetic field always forms a loop that runs from one end of the magnet to the other.
Since this loop of the magnetic field has an origination and termination point which are at the two ends of the magnet (North and South poles). A magnet will always be bipolar which is in this case, North and South; even at an atomic level.
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magnitude 950 \text{ N/C}950 N/C. What is the change in the electric potential energy of the proton-field system when the proton travels to x
Answer:
The change in potential energy is [tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
Explanation:
From the question we are told that
The magnitude of the uniform electric field is [tex]E = 950 \ N/C[/tex]
The distance traveled by the electron is [tex]x = 2.50 \ m[/tex]
Generally the force on this electron is mathematically represented as
[tex]F = qE[/tex]
Where F is the force and q is the charge on the electron which is a constant value of [tex]q = 1.60*10^{-19} \ C[/tex]
Thus
[tex]F = 950 * 1.60 **10^{-19}[/tex]
[tex]F = 1.52 *10^{-16} \ N[/tex]
Generally the work energy theorem can be mathematically represented as
[tex]W = \Delta KE[/tex]
Where W is the workdone on the electron by the Electric field and [tex]\Delta KE[/tex] is the change in kinetic energy
Also workdone on the electron can also be represented as
[tex]W = F* x *cos( \theta )[/tex]
Where [tex]\theta = 0 ^o[/tex] considering that the movement of the electron is along the x-axis
So
[tex]\Delta KE = F * x cos (0)[/tex]
substituting values
[tex]\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)[/tex]
[tex]\Delta KE = 3.8*10^{-16} J[/tex]
Now From the law of energy conservation
[tex]\Delta PE = - \Delta KE[/tex]
Where [tex]\Delta PE[/tex] is the change in potential energy
Thus
[tex]\Delta PE = - 3.8*10^{-16} \ J[/tex]
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor
1) drops to one-fourth its previous value.
2) quadruples.
3) becomes six times its previous value.
4) doubles.
5) drops to one-third its previous value.
6) Not enough information is provided.
7) triples.
8) drops to half its previous value.
9) drops to one-sixth its previous value.
10) remains unchanged.
Answer:
Drop to half of the previous value
Explanation:
Energy stored in capacitor is inversly propotional to the distance between the plates.
If the separation between the capacitor plates is doubled while the capacitor remains connected to the battery, the energy stored in the capacitor drops to half its previous value.
What is parallel plate capacitor?The two parallel plates placed at a distance apart used to store charge when electric supply is on.
The capacitance of a capacitor is given by
C = ε₀ A/d
where, ε₀ is the permittivity of free space, A = area of cross section of plates and d is the distance between them.
Capacitance is inversely proportional to the distance between them. So, if distance is doubled, the capacitance decreases to half its original value.
Thus, the correct option is 8.
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a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
Answer:
q/6Eo
Explanation:
See attached file pls
A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?
a. tension
b. tension and gravity
c. tension, gravity, and the centripetal force
d. tension, gravity, the centripetal force, and friction
Answer:
c. tension, gravity, and the centripetal force
Explanation:
The ball experiences a variety of force as explained below.
Gravity force acts on the body due to its mass and the acceleration due to gravity. The gravity force on every object on earth due to its mass keeps all object on the surface of the earth.
Although the car moves around in circle, centripetal towards the center of the radius of turn exists on the ball. This centripetal force is due to the constantly changing direction of the circular motion, resulting in a force away from the center. The centripetal force keeps the ball from swinging off away from the center of turn.
Tension force on the string holds the ball against falling towards the earth under its own weight, and also from swinging away from the center of turn of the car. Tension force holds the ball relatively fixed in its vertical position in the car.
Suppose that the separation between two speakers A and B is 4.30 m and the speakers are vibrating in-phase. They are playing identical 103-Hz tones and the speed of sound is 343 m/s. An observer is seated at a position directly facing speaker B in such a way that his line of sight extending to B is perpendicular to the imaginary line between A and B. What is the largest possible distance between speaker B and the observer, such that he observes destructive interference
Answer:
The largest possible distance is [tex]x = 4.720 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 4.30 \ m[/tex]
The frequency of the tone played by both speakers is [tex]f = 103 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The wavelength of the tone played by the speaker is mathematically evaluated as
[tex]\lambda = \frac{v}{f}[/tex]
substituting values
[tex]\lambda = \frac{343}{103}[/tex]
[tex]\lambda = 3.33 \ m[/tex]
Let the the position of the observer be O
Given that the line of sight between observer and speaker B is perpendicular to the distance between A and B then
The distance between A and the observer is mathematically evaluated using Pythagoras theorem as follows
[tex]L = \sqrt{d^2 + x^2}[/tex]
Where x is the distance between the observer and B
For the observer to observe destructive interference
[tex]L - x = \frac{\lambda}{2}[/tex]
So
[tex]\sqrt{d^2 + x^2} - x = \frac{\lambda}{2}[/tex]
[tex]\sqrt{d^2 + x^2} = \frac{\lambda}{2} +x[/tex]
[tex]d^2 + x^2 = [\frac{\lambda}{2} +x]^2[/tex]
[tex]d^2 + x^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} + x^2][/tex]
[tex]d^2 = [\frac{\lambda^2}{4} +2 * x * \frac{\lambda}{2} ][/tex]
substituting values
[tex]4.30^2 = [\frac{3.33^2}{4} +2 * x * \frac{3.33}{2} ][/tex]
[tex]x = 4.720 \ m[/tex]
A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. The first reaches the second two and all couple together. The mass of each is 3.05 ✕ 104 kg.(a)What is the speed (in m/s) of the three coupled cars after the first couples with the other two? (Round your answer to at least two decimal places.)Incorrect: Your answer is incorrect.What is the momentum of the two coupled cars? What is the momentum of the first car in terms of its mass and initial speed? Note all cars are initially traveling in the same direction. Apply conservation of momentum to find the final speed. m/s(b)Find the (absolute value of the) amount of kinetic energy (in J) converted to other forms during the collision.J
Answer:
momentum of the coupled cars V = 1.77 m/s
kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J
Explanation:
given
m₁ =3.05 × 10⁴kg
u₁ =2.90m/s
m₂=6.10× 10⁴kg
u₂=1.20m/s
using law of conservation of momentum
m₁u₁ + m₂u₂ = (m₁ + m₂) V
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V
V = 1.617×10⁵/9.15×10⁴
V = 1.77m/s
K.E =1/2mV²
ΔK.E = K.E(final) - K.E(initial)
ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔK.E = -2.892×10⁴J
The final speed is 1.77 m/s
The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]
2.892×10⁴J of energy is converted.
Inelastic collision:Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.
Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg
The initial speed of the railroad boxcar is u₁ = 2.90m/s
Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg
And the initial speed be u₂ = 1.20m/s
The initial momentum of the first car is:
m₁u₁ = 3.05 × 10⁴ × 2.90 = 8.84 × 10⁴ kgm/s
The initial momentum of the coupled car is:
m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s
Let the final speed after all the boxcars are coupled be v
From the law of conservation of momentum, we get:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv
v = 1.617×10⁵/9.15×10⁴
v = 1.77m/s
The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:
ΔKE = K.E(final) - K.E(initial)
ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔKE = -2.892×10⁴J
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a certain plane parallel capacitor stores energy E when the plates have a charge Q on each plate. Then distance between the plates is double. In order to store triply as much energy, how much charge should it have in its plates
Answer:
[tex]Q'=\sqrt{6}Q[/tex]
Explanation:
You have that a parallel plate capacitor has a total energy of E when the distance between the plates is d and the charge on each plate is Q.
You take into account the following formula for the stored energy in the capacitor:
[tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex] (1)
The capacitance C of the parallel plate capacitor is given by the following formula is:
[tex]C=\epsilon_o\frac{A}{d}[/tex] (2)
A: area of the plates
ε0: dielectric permittivity of vacuum
You replace the expression (2) into the equation (1):
[tex]E=\frac{1}{2}\frac{Q^2A}{\epsilon_o d}[/tex] (3)
the previous formula is the expression for the total energy stored for the given parameters A, d and Q.
If the distance between the plates is twice and it is required that the energy is three times the initial energy, to find the value of the charge you use the equation (3):
[tex]E'=\frac{1}{2}\frac{Q'^2A}{\epsilon_o d'}[/tex] (4)
d' = 2d
E' = 3E
Q': required charge
You replace the values of d' and E' in the equation (4) and then divide the result with the equation (3):
[tex]3E=\frac{1}{2}\frac{Q'^2A}{\epsilon_o(2d)}=\frac{1}{4}\frac{Q'^2A}{\epsilon_od}\\\\\frac{3E}{E}=\frac{1/4\frac{Q'^2A}{\epsilon_od}}{1/2\frac{Q^2A}{\epsilon_o d}}\\\\3=\frac{1}{2}\frac{Q'^2}{Q^2}[/tex]
Finally, you solve for Q':
[tex]3=\frac{1}{2}\frac{Q'^2}{Q^2}\\\\Q'=\sqrt{6}Q[/tex]
Then, the required charge is √6Q , to obtain three times the initial energy E, when the distance between plates is doubled.
An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?
Answer:
The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.
Explanation:
The riders will experience a centripetal force from the cylinder
[tex]F_{C}[/tex] = mrω^2 .... equ 1
where
m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of of the rider
For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as
[tex]F_{f}[/tex] = μR ....equ 2
where
μ = coefficient of friction = 0.87
R is the normal force from the rider = mg
where
m is the rider's mass
g is the acceleration due to gravity = 9.81 m/s
substitute mg for R in equ 2, we'll have
[tex]F_{f}[/tex] = μmg ....equ 3
Equating centripetal force of equ 1 and frictional force of equ 3, we'll get
mrω^2 = μmg
the mass of the rider cancels out, and we are left with
rω^2 = μg
ω^2 = μg/r
ω = [tex]\sqrt{\frac{ug}{r} }[/tex]
ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]
ω = 1.58 rad/second
The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s
The riders will experience a centripetal force from the cylinder
[tex]F = mrw^2[/tex]
where m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of the rider
For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as
f = μN
where
μ = coefficient of friction = 0.87
N is the normal force = mg
f = μmg
Equating centripetal force of and frictional force of we'll get
[tex]mrw^2 = umg[/tex]
[tex]rw^2 = ug[/tex]
[tex]w^2 = ug/r[/tex]
[tex]w= \sqrt{ug/r}[/tex]
[tex]w= \sqrt{0.87*9.8/3.4}[/tex]
ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.
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A uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is rotating at an angular velocity of 15 rad/s when a 5-kg package is dropped vertically on a point that is 2.0 m from the center of the disk. What is the angular velocity of the disk/package system
Answer:
18.75 rad/s
Explanation:
Moment of inertia of the disk;
I_d = ½ × m_disk × r²
I_d = ½ × 10 × 4²
I_d = 80 kg.m²
I_package = m_pack × r²
Now,it's at 2m from the centre, thus;
I_package = 5 × 2²
I_package = 20 Kg.m²
From conservation of momentum;
(I_disk + I_package)ω1 = I_disk × ω2
Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.
Thus;
Plugging in the relevant values, we obtain;
(80 + 20)15 = 80 × ω2
1500 = 80ω2
ω2 = 1500/80
ω2 = 18.75 rad/s
Can an object travel at the speed of
light? Why or why nbt?
Answer:
no the only things that can travel at the speed of light are waves in the electromagnetic spectrum
A guitar string 0.65 m long has a tension of 61 N and a mass per unit length of 3.0 g/m. (i) What is the speed of waves on the string when it is plucked? (ii) What is the string's fundamental frequency of vibration when plucked? (iii) At what other frequencies will this string vibrate?
Answer:
i
[tex]v = 142.595 \ m/s[/tex]
ii
[tex]f = 109.69 \ Hz[/tex]
iii1 )
[tex]f_2 =219.4 Hz[/tex]
iii2)
[tex]f_3 =329.1 Hz[/tex]
iii3)
[tex]f_4 =438.8 Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 0.65 \ m[/tex]
The tension on the string is [tex]T = 61 \ N[/tex]
The mass per unit length is [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]
The speed of wave on the string is mathematically represented as
[tex]v = \sqrt{\frac{T}{m} }[/tex]
substituting values
[tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]
[tex]v = 142.595 \ m/s[/tex]
generally the string's frequency is mathematically represented as
[tex]f = \frac{nv}{2l}[/tex]
n = 1 given that the frequency we are to find is the fundamental frequency
So
substituting values
[tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]
[tex]f = 109.69 \ Hz[/tex]
The frequencies at which the string would vibrate include
1 [tex]f_2 = 2 * f[/tex]
Here [tex]f_2[/tex] is know as the second harmonic and the value is
[tex]f_2 = 2 * 109.69[/tex]
[tex]f_2 =219.4 Hz[/tex]
2
[tex]f_3 = 3 * f[/tex]
Here [tex]f_3[/tex] is know as the third harmonic and the value is
[tex]f_3 = 3 * 109.69[/tex]
[tex]f_3 =329.1 Hz[/tex]
3
[tex]f_3 = 4 * f[/tex]
Here [tex]f_4[/tex] is know as the fourth harmonic and the value is
[tex]f_3 = 4 * 109.69[/tex]
[tex]f_4 =438.8 Hz[/tex]
An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?
Answer:
Towards the west.
Explanation:
The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.
Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.
According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward
Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come
Two vehicles approach an intersection, a 2500kg pickup travels from E to W at 14.0m/s and a 1500kg car from S to N at 23.0m/s. Find P net of this system (direction and magnitude)
Answer:
The magnitude of the momentum is 49145.19 kg.m/s
The direction of the two vehicles is 44.6° North West
Explanation:
Given;
speed of first vehicle, v₁ = 14 m/s (East to west)
mass of first vehicle, m₁ = 1500 kg
speed of second vehicle, v₂ = 23 m/s (South to North)
momentum of the first vehicle in x-direction (E to W is in negative x-direction)
[tex]P_x = mv_x\\\\P_x = 2500kg(-14 \ m/s)\\\\P_x = -35000 \ kg.m/s[/tex]
momentum of the second vehicle in y-direction (S to N is in positive y-direction)
[tex]P_y = m_2v_y\\\\P_y = 1500kg(23 \ m/s)\\\\P_y = 34500 \ kg.m/s[/tex]
Magnitude of the momentum of the system;
[tex]P= \sqrt{P_x^2 + P_y^2} \\\\P = \sqrt{(-35000)^2+(34500)^2} \\\\P = 49145.19 \ kg.m/s[/tex]
Direction of the two vehicles;
[tex]tan \ \theta = \frac{P_y}{|P_x|} \\\\tan \ \theta = \frac{34500}{35000} \\\\tan \ \theta = 0.9857\\\\\theta = tan^{-1} (0.9857)\\\\\theta = 44.6^0[/tex]North West
You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current in the wire is 1.2A. What is the radius of the coils and how many loops, N, are there
Answer:
radius of the loop = 7.9 mm
number of turns N ≅ 399 turns
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = [tex]4\pi *10^{-7} T-m/A[/tex] = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop = [tex]2\pi r[/tex]
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = [tex]7.9*10^{-4} m[/tex] = 0.0079 m = 7.9 mm
Light of wavelength λ=0.01nm, is scattered at 1350 from a stationary electron. What is the kinetic energy of the recoiling electron?
Answer:
198.9 x 10^-16
Explanation:
E = hc/ wavelength
E =(6.63 x 10^-34 x 3 x 10^8)/(0.01 x 10^-9)
E = 198.9 x 10^-16