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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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 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
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
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.
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 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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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
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
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 friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
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
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 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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how far do you think you would go in a car while sneezing for 2.5 seconds
Answer: If you are traveling at a speed of 60mph, you will go 220 feet.
Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.
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 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]
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:
As you drive down the road at 13 m/s , you press on the gas pedal and speed up with a uniform acceleration of 1.02 m/s2 for 0.70 s. If the tires on your car have a radius of 33 cm, what is their angular displacement during this period of acceleration?
Answer:
The angular displacement is [tex]\theta = 28.33 \ rad[/tex]
Explanation:
From the question we are told that
The speed of the driver is [tex]v =13 \ m/ s[/tex]
The acceleration of the driver is [tex]a = 1.02 \ m/s^2[/tex]
The time taken is [tex]t = 0.70 \ s[/tex]
The radius of the tire is [tex]r = 33 cm = 0.33 \ m[/tex]
The distance covered by the car during this acceleration can be calculated using the equation of motion as follows
[tex]s = v*t +\frac{1}{2} * a * t^2[/tex]
Now substituting values
[tex]s = 13 * 0.70 +\frac{1}{2} * 1.02 * (0.700)^2[/tex]
[tex]s = 9.35 \ m[/tex]
Now the angular displacement of the car with respect to the tire movement can be represented mathematically as
[tex]\theta = \frac{s}{r}[/tex]
substituting values
[tex]\theta = \frac{9.35}{0.33}[/tex]
[tex]\theta = 28.33 \ rad[/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
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
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.
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
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
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
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.
Learn more about parallel plate capacitor.
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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]
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.
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