Answer:
Normal force = 3N
Explanation:
We are given;
Coefficient of kinetic friction friction; μ_k = 0.3
Frictional force;F_f = 0.1N
Now,the formula for frictional force is;
F_f = μ_k*N
Where μ_k is coefficient of friction and N is the normal force.
So, making N the subject, we have;
N = μ_k/F_f
Plugging in the relevant values to obtain;
N = 0.3/0.1
N = 3N
A wire with length 3.5 m and mass 0.03 kg is aligned horizontally and connected by conducting loops to two vertical frictionless metal rods. The rod is oriented in the East-West direction. What is the magnitude of the current that has to be run through the wire so that it hovers in place without sliding up or down? You may ignore magnetic forces caused by the vertical wires. Earth's magnetic field: 0.00005 Tesla or 0.5 gauss. Earth's gravity: 9.8 N/kg
Answer:
Current flowing in the wire is 1680 A
Explanation:
It is given length of wire l = 3.5 m
Mass of the wire m = 0.03 kg
Magnetic field B = 0.00005 Tesla
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Mg force acting on the wire will be equal to Lorentz force acting on the wire.
Therefore [tex]mg=IBl[/tex]
[tex]0.03\times 9.8=I\times 0.00005\times 3.5[/tex]
[tex]I=1680A[/tex]
Therefore current flowing in the wire is 1680 A
Power of sunlight on Earth The Sun emits about 3.9 * 1026 J of electromagnetic radiation each second. (a) Estimate the power that each square meter of the Sun’s surface radiates. (b) Estimate the power that 1 m2 of Earth’s surface receives. (c) What assumptions did you make in part (b)? The distance from Earth to the Sun is about 1.5 * 1011 m and the diameter of the Sun is about 1.4 * 109 m
Answer:
a)6.34 x [tex]10^{7}[/tex]W/m²
b)1.37 x [tex]10^{3[/tex] W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x [tex]10^{26}[/tex] J
and the area of the sun is given by,
A= 4π[tex]R_{sun}[/tex] => 4π[tex](\frac{1.4*10^{9} }{2} )^{2}[/tex]
A=6.15 x [tex]10^{18}[/tex]m²
[tex]I_{sun}[/tex] = 3.9 x [tex]10^{26}[/tex] / 6.15 x [tex]10^{18}[/tex] => 6.34 x [tex]10^{7}[/tex]W/m²
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π[tex]R[/tex]
Now R= 1.5 x [tex]10^{11[/tex]m
A= 4π x 1.5 x [tex]10^{11[/tex] =>2.83 x [tex]10^{23[/tex] m²
The power that each square meter of Earths surface receives
[tex]I_{earth[/tex] = 3.9 x [tex]10^{26}[/tex]/2.83 x [tex]10^{23[/tex] =>1.37 x [tex]10^{3[/tex] W/m²
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.
The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The maximum power that reaches Earth's surface is approximately 1.30 kW/m². The assumptions made in this estimation include the perfect absorption of sunlight by the Earth's surface and the average radius of the Earth.
Explanation:The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The power output of the Sun is given as 4.00 × 10^26 W. To estimate the power reaching Earth's surface, we need to take into account that part of the Sun's radiation is absorbed and reflected by the atmosphere. The maximum power that reaches Earth's surface is approximately 1.30 kW/m².
To estimate the power received by 1 m² of Earth's surface, we can use the fact that the area of the Earth facing the Sun is πR², where R is the radius of the Earth. Assuming the Earth is a perfect sphere, we can take the average radius of the Earth as Re ≈ 6,371 km. Therefore, the power received by 1 m² of Earth's surface can be calculated as (1.30 kW/m²) × (πR²) / (4πRe²), where Re is the radius of the Earth.
In part (b), the assumptions made include the perfect absorption of sunlight by the Earth's surface, the average radius of the Earth, and the assumption that the Earth is a perfect sphere.
Summer temperatures are usually warmer than
A. winter temperatures.
B. spring temperatures.
C. autumn temperatures.
D. all of these
D _____________________________________________________
A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.2 kg block is projected up the plane, from an initial position that is distance d = 0.90 m from the end of the relaxed spring, with an initial kinetic energy of 29 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.30 m? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.40 m?
Answer:
Explanation:
Initial kinetic energy = 29 J
work done against gravity = mgsin33 x d , m is mass of the block
= 1.2 x 9.8 sin 33 x .9
= 5.76 J
potential energy stored in compressed spring
= 1/2 k x², k is spring constant and x is compression
= .5 x 200 x .3²
= 9
energy left = 29 - ( 5.76 + 9 )
= 14.24 J
b )
energy stored in spring when compression is .4 m
= 1/2 x 200 x .4²
= 16 J
required kinetic energy = 16 + 5.76
= 21.76 J
Block must be projected with energy of 21.76 J .
3. A hydraulic lift is used to jack a 950-kg car 45 cm off the floor. The diameter of the output piston is 23 cm, and the input force is 375 N. a. What is the area of the input piston? (10pts) b. What is the work done in lifting the car 45 cm? (10pts) c. If the input piston moves 15 cm in each stroke, how high does the car move up for each stroke? (10pts) d. How many strokes are required to jack the car up 45 cm? (10pts) e. Show that the energy is conserved. (10pts)
Answer:
a. A input = 0.001669 m²
b. W= 4.193775 KJ
c. h output = 0.60357 cm
d. n = 74.55638 strokes
e. 4.1937 N = 4.1937 N
Explanation:
Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.
Parameters given:
Diameter of the output piston, d = 0.23 m
Mass of the car, m= 950 kg
Force applied at the input piston, f = 375 N
Height, h = 0.45 m
(a) Finding the area of the input piston:
First, we use Pascal's principle to find the area
(f ÷ A output) ÷ (f ÷ A input)
Where A= area
g = 9.81
A output = πd² ÷ 4
(f ÷ (πd² ÷ 4)) = (f ÷ A input)
[(950 x 9.81) ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input
9319.5 ÷ 0.0415 = 375 ÷ A input
A input = 0.001669 m²
(b) Finding the work done in lifting the car 45 cm
Work done, W = force, f x distance ( which in this case is height, h)
= (950 x 9.81) x 0.45
W= 4.193775 KJ
(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.
W output = W input
F x h output = F x h input
= (950 x 9.81) x h output = 375 x 0.15
h output = 0.0060357 m
h output = 0.60357 cm
(d) Finding the number of strokes that are required to jack the car up 45 cm
n = h ÷ h output
n = 45 ÷ 0.60357 cm
n = 74.55638 strokes
(e) How the energy is conserved
W output = W input
F x h output = F x h input x n
(950 x 9.81) x 0.45 = 375 x 0.15 x 74.55638
4.1937 N = 4.1937 N
The area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
Given information:
Mass of the car is [tex]m=950[/tex] kg
The lift of the car is [tex]h=45[/tex] cm.
The diameter of the output piston D is 23 cm.
The input force f is 375 N.
Let the output force be F and the input piston diameter be d.
Hydraulic lift follows Pascal's principle.
(a)
So, the area of the input piston will be calculated as,
[tex]\dfrac{F}{\frac{\pi}{4}D^2}=\dfrac{f}{\frac{\pi}{4}d^2}\\\dfrac{\pi}{4}d^2=\dfrac{f\frac{\pi}{4}D^2}{mg}\\\dfrac{\pi}{4}d^2=A=0.00167\rm\;m^2[/tex]
(b)
The work done in lifting the car will be,
[tex]W=mgh\\W=950\times 9.81\times0.45\\W=4193.78\rm\;J[/tex]
(c)
The input piston moves 15 cm in each stroke.
The height h' moved by car in each stroke will be calculated as,
[tex]mgh'=f\times 0.15\\h'=0.006035\rm\;m=0.6035\rm\;cm[/tex]
(d)
The number of strokes required to lift the car by 45 cm will be,
[tex]n=\dfrac{h}{h'}\\n=\dfrac{45}{0.6035}\\n=74.57[/tex]
(e)
The energy is conserved when output work is equal to input work.
So, the energy conservation can be verified as,
[tex]W_i=W_o\\f\times0.15\times n=mgh\\375\times 0.15\times74.57=950\times 9.81\times0.45\\4194.28\approx4194.1[/tex]
The input and output works are approximately equal. The value isn't exactly equal because of the rounding-off.
So, energy conservation is justified.
Therefore, the area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.
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A 10.0-g bullet is fired into a stationary block of wood having mass 5.00 kg. The bullet embeds 10 pts into the block and the speed of the block-and-bullet after the collision is 0.600 m/s. Find a) the original speed of the bullet, b) the mechanical energies of the block-bullet system before and after the collision, c) the percentage of mechanical energy lost to heat.
Answer:
A. The initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J
C. Percentage of K.E lost to heat is = 99.8 %
Explanation:
From conservation of linear momentum,
[tex](m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v[/tex]
let the mass of the block be m1 and velocity = v1
let the mass of the bullet be m2 and velocity = v2
Let the final velocity of the system be v.
A. Plugging our parameters into the equation, we have:
[tex][(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6[/tex]
[tex]v_{2}=\frac{3.006}{0.01}= 300.6m/s[/tex]
Hence, the initial velocity of the bullet is [tex]= 300.6m/s[/tex]
B. The mechanical energies of the system exist in form of kinetic energy.
I. Kinetic energy of the system before collision:
[tex]0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J[/tex]
II. Kinetic energy after collision:
[tex]0.5\times 5.01 \times 0.6^{2}= 0.9018 J[/tex]
C. Change in Mechanical Energy = [tex]451.8 - 0.9018 J= 450.9J[/tex]
[tex]\frac{450.9}{451.8} \times 100 =99.8%[/tex]
Percentage of K.E lost to heat is = 99.8 %
One propulsion method that is currently being studied for interstellar travel uses a large sail. The intent is that sunlight striking the sail creates a force that pushes the ship away from the sun, much as wind propels a sailboat. Does such a design have any hope of working and, if so, should the surface facing the sun be shiny like a mirror or black, in order to produce the greatest force
Answer:
Yes, the design has a hope of working.
A dull face should yield more force.
Explanation:
According to the de broglie theory, and the wave-particle duality, waves can be assumed and made to behave as a matter and vice versa. Under these circumstances, the impinging wave can act as a particle (photon) colliding with a relatively stationary body, yielding some of its momentum to the relatively stationary body. With this, the momentum of the photons can be used to propel the ship by its solar sail.
Making the face dull should yield more force in the sense that the incident photon is not reflected and all of its momentum is yielded to the sail compared to when it is reflected (bouncing off) off the sail. This will be a case of inelestic collision between the sail and the photon.
Once you’ve identified the initial and final conditions, you’re ready to solve for the unknown quantity in your problem. Boyle’s Law expresses the pressure-volume relationship as P1V1=P2V2, so you will need to solve this equation for the unknown quantity and then plug in your known values to calculate the unknown. What pressure would it take to compress 250. L of helium gas initially at 1.00 atm into a 2.00 L tank at constant temperature? Express your answer with the appropriate units.
Answer:
125atm
Explanation:
Given:
initial pressure [tex]P_{1[/tex]= 1 atm
initial volume [tex]V_{1[/tex]= 250L
final volume [tex]V_{2[/tex]=2L
Required:
Final Pressure [tex]P_{2[/tex]=?
As Boyle’s Law expresses the pressure-volume relationship as
[tex]P_{1[/tex][tex]V_{1[/tex]= [tex]P_{2[/tex] [tex]V_{2[/tex]
(1)(250)= [tex]P_{2[/tex] (2)
[tex]P_{2[/tex] = 250/2
[tex]P_{2[/tex]=125atm
Therefore, it would take pressure of 125atm.
Using Boyle's Law, the pressure required to compress a gas of 250 L at 1 atm into a smaller volume of 2 L at constant temperature is calculated to be 125 atm.
Explanation:The student is attempting to solve this problem using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. In this context, the formula for Boyle's Law is P1V1 = P2V2.
Here, the initial conditions are a volume V1 of 250 L and a pressure P1 of 1.00 atm. The final volume V2 is given as 2.00 L, and we are asked to solve for the final pressure P2.
Replace the given values into the Boyle's Law equation: (1.00 atm) * (250.L) = P2 * (2.00 L). From this, we can solve for P2 which sums to: P2 = ((1.00 atm) * (250.L)) / (2.00 L) = 125 atm. The pressure required to compress the helium gas into a 2L container would be 125 atm.
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What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 1.5 times that of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c) as a fraction of the wavelength.
Answer:
a) 82.8°
b)1.44 rad
c)0.23λ
Explanation:
Wave function form for each wave will be
-> wave 1:
[tex]y_{1}[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt)
-> wave 2:
[tex]y_{2[/tex](x,t)=[tex]y_{m[/tex]sin( kx - ωt +φ)
The summation of above two functions is the resultant wave.
Y(x,t)=[tex]y_{m[/tex]sin( kx - ωt)+[tex]y_{m[/tex]sin( kx - ωt +φ)
Using the trigonometric addition formula for sin i.e
sin(A)+ sin(B) = 2 sin(A+B/2) cos(A-B/2)
Y(x,t)=2[tex]y_{m[/tex]cos(φ/2) sin(kx - ωt +φ)
When comparing to the general wave form, the amplitude is 2[tex]y_{m[/tex]cos(φ/2)m
Also, [tex]y_{m[/tex]cos(φ/2)= 1.5 [tex]y_{m[/tex]
φ/2=[tex]cos^{-1}[/tex](1.5/s)
(a) φ= 82.8°
(b)φ= 82.8π/180 =>1.44 rad
(c) one complete wave is 2π radians
Therefore, for two waves, φ/2π= 1.44/2π => 0.23 fraction of a complete wave from each other i.e they are separated by 0.23λ
A red laser from a physics lab is marked as producing 632.8 nm light. When light from this laser falls on two closely spaced slits, and interference pattern formed on a wall several meters away has bright fringes spaced 5.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 5.14 mm apart. What is the wavelength of the light produced by the laser pointer?
Given Information:
Wavelength of the red laser = λr = 632.8 nm
Distance between bright fringes due to red laser = yr = 5 mm
Distance between bright fringes due to laser pointer = yp = 5.14 mm
Required Information:
Wavelength of the laser pointer = λp = ?
Answer:
Wavelength of the laser pointer = λp = ?
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
y = Dλ/d
y/λ = D/d
Where
λ is the wavelength
y is the distance between bright fringes.
d is the double slit separation distance
D is the distance from the slits to the screen
For the red laser,
yr/λr = D/d
For the laser pointer,
yp/λp = D/d
Equating both equations yields,
yr/λr = yp/λp
Re-arrange for λp
λp = yp*λr/yr
λp = (5*632.8)/5.14
λp = 615.56 nm
Therefore, the wavelength of the small laser pointer is 615.56 nm.
Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so that the magnetic field of one cannot affect the other. (a) Calculate the equivalent inductance. (Hint: Review the derivations for resistors in parallel and capacitors in parallel.Which is similar here?) (b) What is the generalization of (a) for N = 10 similar inductors L = 3.22 H in parallel?
Answer:
a) 0.83H
b) 3.22/N Henry
Explanation:
Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.
Derivation:
If the voltage across an inductor
VL = IXL
I is the current
XL is the inductive reactance
XL = 2πfL
VL = I(2πfL)
L is the inductance.
From the formula, I = V/2πfL
Given two inductors in parallel, different current will flow through them.
I1 = V/2πfL1 (current in L1)
I2 = V/2πfL2 (current in L2)
Total current I = I1+I2
I = V/2πfL1 + V/2πfL2
I = V/2πf{1/L1+1/L2}
V/2πfL = V/2πf{1/L1+1/L2}
1/L = 1/L1+1/L2 (equivalent inductance in parallel)
Given L1 = 1.31 H and L2 = 2.24
1/L = 1/1.31 + 1/2.24
1/L = 0.763 + 0.446
1/L = 1.209
L = 1/1.209
L = 0.83H
The equivalent inductance is 0.83H
b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:
1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22
+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22
1/L = 10/3.22 (since that all have the same denominator)
L = 3.22/10
If N = 10, the generalization of 10 similar inductors in parallel will be:
L = 3.22/N Henry
A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and? a. Find the tension in the string. (10pts) b. What is the new balance reading? (10pts)
Answer:
a
The tension in the string is [tex]T = 0.85 N[/tex]
b
The new balance reading is [tex]M_b = 885.86 g[/tex]
Explanation:
From the question we are told that
The mass of the beaker of water is [tex]m = 875 .0g[/tex]
The diameter of the copper ball is [tex]d = 2.75 cm = \frac{2.75}{100} = 0.0275 m[/tex]
There are two forces acting on the copper ball
The first is the Buoyant force of the water pushing it up which is mathematically represented as
[tex]F = \rho V g[/tex]
Where [tex]\rho[/tex] is the density of water which has value of [tex]\rho = 1000 kg/m^3[/tex]
g is the acceleration due to gravity [tex]g= 9.8 \ m/s^2[/tex]
[tex]V[/tex] is the volume of water displaced by the copper ball which is mathematically evaluated as
[tex]V = \frac{4}{3} \pi r^3[/tex]
The radius r is [tex]r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m[/tex]
Substituting value
[tex]V = \frac{4}{3} * 3.142 * (0.01375)^3[/tex]
[tex]V = 1.08 9 *10^{-5 } m^3[/tex]
Substituting for F
[tex]F = 1000 * 1.089 *10^{-5} * 9.8[/tex]
[tex]F = 0.1067 N[/tex]
The second force is the weight of the copper ball which is mathematically represented as
[tex]W_c = mg[/tex]
Now m is the mass which can be mathematically evaluated as
[tex]m = \rho_c * V[/tex]
Where is the density of copper with value of [tex]\rho_c = 8960 kg /m^3[/tex]
So [tex]m = 8960 * 1.089 *10^{-5}[/tex]
[tex]m = 0.0976[/tex]
So the weight of copper is
[tex]W_c = 0.09756 * 9.8[/tex]
[tex]W_c = 0.956 N[/tex]
Now the tension the string would be mathematically evaluated as
[tex]T = W_c - F[/tex]
So [tex]T = 0.956 - 0.1067[/tex]
[tex]T = 0.85 N[/tex]
From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance
Now the mass equivalent of this weight is mathematically evaluated as
[tex]m_z = \frac{1.0645 }{9.8 }[/tex]
[tex]m_z = 0.01086 kg[/tex]
Converting to grams
[tex]m_z = 10.86 g[/tex]
So the new balance reading is
[tex]M_b = 875.0 +10.86[/tex]
[tex]M_b = 885.86 g[/tex]
6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s, radius 10.0 km, and mass 2.8 × 1030 kg. The pulsar’s rotational period will increase over time due to the release of electromagnetic radiation, which doesn’t change its radius but reduces its rotational energy. (a) What is the angular momentum of the pulsar? (b) Suppose the angular velocity decreases at a rate of 10−14 rad/s2 . What is the torque on the pulsar?
Answer:
a) L = 2.10x10⁴⁰ kg*m²/s
b) τ = 1.12x10²⁴ N.m
Explanation:
a) The angular momentum (L) of the pulsar can be calculated using the following equation:
[tex] L = I \omega [/tex]
Where:
I: inertia momentum
ω: angular velocity
First we need to calculate ω and I. The angular velocity can be calculated as follows:
[tex] \omega = \frac{2 \pi}{T} [/tex]
Where:
T: is the period = 33.5x10⁻³ s
[tex] \omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s [/tex]
The inertia moment of the pulsar can be calculated using the following relation:
[tex] I = \frac{2}{5}mr^{2} [/tex]
Where:
m: is the mass of the pulsar = 2.8x10³⁰ kg
r: is the radius = 10.0 km
[tex] I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2} [/tex]
Now, the angular momentum of the pulsar is:
[tex] L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1} [/tex]
b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:
[tex] \tau = I*\alpha [/tex]
Where:
α: is the angular acceleration = 10⁻¹⁴ rad/s²
[tex]\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m[/tex]
I hope it helps you!
Final answer:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Explanation:
To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid sphere can be calculated using the formula I = (2/5)mr², where m is the mass and r is the radius.
Using the given radius and mass of the pulsar, we can calculate the moment of inertia.
Once we have the moment of inertia, we can use the given period of the pulsar to calculate the angular velocity.
To find the torque on the pulsar, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
The angular acceleration can be calculated using the given rate at which the angular velocity is decreasing.
Using these formulas and the given values, we can find the angular momentum and torque of the pulsar.
A musician on a unicycle is riding at a steady 6.2 m/s while playing their well tuned oboe at an A above middle C (440 Hz). They then ride toward another musician sitting on a park bench also playing an A above middle C. However it appears to them that one of them is off as they hear beats. If we assume both are tuned properly and the outside temperature is 27 oC what will the beat frequency they hear be
Answer:
The frequency of the beat is [tex]f_t = 8Hz[/tex]
Explanation:
From the question we are told that
The speed of the ride [tex]v = 6,2 m/s[/tex]
The frequency of the oboe is [tex]f = 440Hz[/tex]
The temperature outside is [tex]T = 27^oC[/tex]
Generally the speed of sound generated in air is mathematically evaluated as
[tex]v_s = 331 + 0.61 T[/tex]
Substituting value
[tex]v_s = 331 + 0.61 *27[/tex]
[tex]v_s = 347.47 \ m/s[/tex]
The frequency of sound(generated by the musician on he park) getting to the musicians on the unicycle is mathematically evaluated as
[tex]f_a = \frac{v_s }{v_s - v} f[/tex]
substituting values
[tex]f_a = \frac{347.47 }{347.47 - 6.2} * 440[/tex]
[tex]f_a = 448Hz[/tex]
Since the musician on the park is not moving the frequency of sound (from the musicians riding the unicycle )getting to him is = 440Hz
The beat frequency these musician here is mathematically evaluated as
[tex]f_t = f_a - f[/tex]
So [tex]f_t = 448 - 440[/tex]
[tex]f_t = 8Hz[/tex]
A satellite and the International Space Station have the same mass and are going around Earth in concentric orbits. The distance of the satellite from Earth\'s center is twice that of the International Space Station\'s distance. What is the ratio of the centripetal force acting on the satellite compared to that acting on the International Space Station
Answer:
[tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
Explanation:
G = Gravitational constant
r = Distance between Earth and object
M = Mass of Earth
m = Mass of object
Centripetal force on the space station
[tex]F_1=\dfrac{GMm}{r^2}[/tex]
Centripetal force on the satellite
[tex]F_2=\dfrac{GMm}{(2r)^2}\\\Rightarrow F_2=\dfrac{GMm}{4r^2}[/tex]
From the question the required ratio is
[tex]\dfrac{F_2}{F_1}=\dfrac{\dfrac{GMm}{4r^2}}{\dfrac{GMm}{r^2}}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
The ratio is [tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]
Dan picks up a 15-m-long pole and begins running very fast, holding the pole horizontally and pointing in the direction he's running. He heads toward a barn that is12 m long and has open doors at each end. Dan runs so fast that, to Farmer Brown standing by his barn, the ladder is only 5 m long. As soon as the pole is completely inside the barn, Farmer Brown closes both doors so that Dan and the pole are inside with both doors shut. Then, just before Dan reaches the far door, Farmer Brown opens both doors and Dan emerges, still moving at high speed. According to Dan, however, the barn is contracted to only 4 m and the pole has its full 15 m length. Farmer Brown sees the pole completely inside the barn with both doors closed.
In the farmer's reference frame, eight events occur in the following order:
1. The front of the pole reaches the back of the barn (FB).
2. The back of the pole reaches the back of the barn (BB).
3, 4. (simultaneous events) The farmer shuts the front door (SF); the farmer shuts the back door (SB).
5. The front of the pole reaches the front of the barn (FF).
6, 7. (simultaneous events) The farmer opens the front door (OF). The farmer opens the back door (OB).
8. The back of the pole reaches the front of the barn (BF).
What does Dan see happening (in the chronological order)?
In the farmer's reference frame, eight events occur in the following order:
1. The front of the pole reaches the back of the barn (FB).
2. The back of the pole reaches the back of the barn (BB).
3, 4. (simultaneous events) The farmer shuts the front door (SF); the farmer shuts the back door (SB).
5. The front of the pole reaches the front of the barn (FF).
6, 7. (simultaneous events) The farmer opens the front door (OF). The farmer opens the back door (OB).
8. The back of the pole reaches the front of the barn (BF).
What does Dan see happening (in the chronological order)?
1. FB; 2, 3. FF and OF are simultaneous; 4. SF; 5, 6. BB and OB are simultaneous; 7. SB; 8 BF.
1. FB; 2. FF; 3, 4. SF and OF are simultaneous; 5, 6. OB and SB are simultaneous; 7. BB; 8 BF.
1. FB; 2, 3. SF and OF are simultaneous; 4. FF; 5, 6. OB and SB are simultaneous; 7. BB; 8 BF.
1. FB; 2. SF; 3, 4. FF and OF are simultaneous; 5, 6. BB and SB are simultaneous; 7. OB; 8 BF.
Answer:
1. FB; 2. SF; 3, 4. FF and OF are simultaneous; 5, 6. BB and SB are simultaneous; 7. OB; 8 BF
Explanation:
According to the report the listed below are what Dan saw in the exact chronological order.
1. FB; 2. SF; 3, 4. FF and OF are simultaneous; 5, 6. BB and SB are simultaneous; 7. OB; 8 BF
These represents the series of events as described in the question.
Which number multplied by nine is equal to 81?
O 7
O 8
O9
Answer:
9
Explanation:
9×9 is 81, i hope this helpes
Answer:
9
Explanation:
A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?
Given Information:
Resistance = R = 14 Ω
Inductance = L = 2.3 H
voltage = V = 100 V
time = t = 0.13 s
Required Information:
(a) energy is being stored in the magnetic field
(b) thermal energy is appearing in the resistance
(c) energy is being delivered by the battery?
Answer:
(a) energy is being stored in the magnetic field ≈ 219 watts
(b) thermal energy is appearing in the resistance ≈ 267 watts
(c) energy is being delivered by the battery ≈ 481 watts
Explanation:
The energy stored in the inductor is given by
[tex]U = \frac{1}{2} Li^{2}[/tex]
The rate at which the energy is being stored in the inductor is given by
[tex]\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1[/tex]
The current through the RL circuit is given by
[tex]i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })[/tex]
Where τ is the the time constant and is given by
[tex]\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16[/tex]
[tex]i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}[/tex]
Therefore, eq. 1 becomes
[tex]\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})[/tex]
At t = 0.13 seconds
[tex]\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts[/tex]
(b) thermal energy is appearing in the resistance
The thermal energy is given by
[tex]P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts[/tex]
(c) energy is being delivered by the battery?
The energy delivered by battery is
[tex]P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts[/tex]
Type A.
1. (a) A rectangular frame of wire of 25 cm x 15 cm is placed
in a uniform electric field of strength 2 x 10-NC-, such
that the plane of the coil is normal to field. Find the electric
flux linked with the rectangular frame.
(6) Calculate the electric flux linked with the frame, when
it is converted into (1) a square and (ii) a circular frame.
(c) In which case is the electric flux maximum ?
O m ? C-1 (h) () SON m2 -1.
Answer:
a) 750W
b) 800W
1018.9W
c) the circular frame
Explanation:
a) To find the electric flux you use the following formula:
[tex]\Phi_E=EAcos\theta=EA[/tex]
where you have assumed that the plane of the loop and B are perpendicular between them.
By replacing you obtain:
[tex]\Phi_E=(2*10^4N/C)(0.25m)(0.15m)=750W[/tex]
b) In order to calculate the electric flux in a square frame you first calculate the perimeter of the rectangular frame of wire:
P = 0.25m+0.25m+0.15m+0.15m=0.8m
A square with this length will have sides of 0.2m
Hence. you have that the electric flux is:
[tex]\Phi_E=(2*10^4N/C)(0.2m)^2=800W[/tex]
for a circular frame you have that the radius is:
[tex]s=2\pi r\\\\r=\frac{s}{2\pi}=\frac{0.8m}{2\pi}=0.127m[/tex]
Then, the electric flux will be:
[tex]\Phi_E=(2*10^4N/C)(\pi)(0.127m)^2=1018.9W[/tex]
c) The electric flux is maximum in the circular frame.
Diesel engines give more miles per gallon than gasoline engines, but some of this is due to the higher energy content of diesel fuel. At highway speeds, it takes 0.20 MJ to move an aerodynamic car 1.0 km. At highway speeds, with a gasoline engine, a car gets 16 km per liter of fuel; with a diesel engine under the hood, the car gets 19 km per liter of fuel. One liter of diesel contains 36 MJ; 1 liter of gasoline contains only 32 MJ. Part A What is the efficiency of the diesel engine
Answer:
0.106
Explanation:
For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:
[tex]\frac{3.8}{36} = 0.106[/tex]
The efficiency of a diesel engine, in the given scenario, is found to be 10.56%. This is determined by considering the energy content of diesel fuel and the amount of work done by the engine.
Explanation:Firstly, the energy content for a liter of diesel fuel is 36 MJ, and the car can travel 19 km with one liter of fuel. Thus, the total energy to travel 19 km is 19 km * 0.20 MJ/km = 3.8 MJ. The useful outcome is therefore 3.8 MJ from the total of 36 MJ. Therefore, applying the formula for efficiency which is Output Energy/Input Energy * 100%, gives a result of (3.8 MJ / 36 MJ) * 100% = 10.56%.
This indicates that the diesel engine is 10.56% efficient at converting the energy in diesel fuel into work to move the car, whereas the remaining energy is lost mainly due to heat, sound etc.
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A ship is floating on a lake. Its hold is the interior space beneath its deck; the hold is empty and is open to the atmosphere. The hull has a hole in it, which is below the water line, so water leaks into the hold. The effective area of the hole is 4.8 × 10-3 m2 and is located 2.5 m beneath the surface of the lake. What volume of water per second leaks into the ship?
Answer:
33.4 litres/sec
Explanation:
Given:
hole area A = 0.0048 m²
Velocity is given by:
V =√2gh = √(2 x 9.8 x 2.5) = 7 m/sec
flow = A x V = (0.0048)(7) = 0.0334 m³/sec
0.0334 m³/sec => 33.4 litres/sec
No water leaks into the ship because the velocity of the water entering the hole is 0 m/s.
Explanation:The volume of water leaking into the ship can be calculated using Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The volume of water displaced is equal to the volume of water leaking into the ship. The formula to calculate the volume of water leaking per second is: V = A * h * v, where V is the volume, A is the area of the hole, h is the height of the water above the hole, and v is the velocity of the water entering the hole. In this case, the area of the hole is given as 4.8 × 10-3 m2, the height of the water above the hole is 2.5 m, and we can assume the velocity of the water entering the hole is 0 m/s since the ship is floating on a lake and there is no external force pushing the water into the hole.
Using the formula, we can calculate the volume of water leaking per second:
V = (4.8 × 10-3 m2) * (2.5 m) * 0 m/s = 0 m3/s
Therefore, no water is leaking into the ship because the velocity of the water entering the hole is 0 m/s.
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A current flowing through you of more than 5 ma is considered dangerous. Why do we see warnings about high voltage, rather than high current? What voltages are dangerous even if the skin is dry? (Dry skin has a resistance of 100 kohms while wet skin can be much less.) An electroencephalogram (EEG) is a recording of differences in voltage observed on the scalp by attaching electrical contacts. If voltage differences that are observed are about 0.5 mv, and the resistance of the skin is about 10 kohms, what size currents are involved?
Answer:
V = 500 volts
I = 50 nA
Explanation:
We usually see 'high voltage' sign rather than 'high current' this is due to the fact that the amount of current depends upon the resistance of the body,
The relation of voltage, current and resistance is given by Ohm's law,
V = IR
I = V/R
As you can see, current and resistance are inversely related, the lower the resistance of the body, the higher will be the current flowing through the body.
What voltages are dangerous even if the skin is dry?
A dry human body has a resistance of approximately 100 kΩ and it gets reduced if the body is wet.
For a current of 5 mA, the corresponding voltage is,
V = 0.005*100,000
V = 500 volts
Therefore, for a dry human body, a voltage of 500 volts would be dangerous.
If voltage differences that are observed are about 0.5 mV, and the resistance of the skin is about 10 kΩ, what size currents are involved?
I = V/R
I = 0.0005/10,000
I = 50×10⁻⁹ A
I = 50 nA
Therefore, a current of 50 nA would be flowing through the EEG.
An object is moving along a line with velocity v (t )equals A t squared space plus space B meters per second at time t seconds. Its initial position, at time t equals 0, is s (0 )equals C meters. If A equals 15, B equals 8 and C equals space 3, then Give the object's position function s (t ). Give the distance traveled by the object between t equals 1 and t equals 2 seconds and record your answer below.
Answer:
- S(t) = 5t³+8t+3
- 19.7m approximately
Explanation:
Given the velocity if an object to be
v(t) = At² + B where t is the time in seconds
Velocity is the change in displacement of a body with respect to time.
V(t) = dS(t)/dt
Making S(t) the subject of the formula
dS(t) = v(t)dt
Integrating both sides
∫dS(t) =∫v(t)dt
S(t) = ∫(At²+B)dt
S(t) = At³/3+Bt + C... 1
In its initial position s(0) = 0
t = 0, s = 0
S(0) = A(0)³/3+B(0)+C
S(0)= C
a) The object's position function s(t) if A = 15, B = 8 and C = 3 can be gotten by substituting this value into eqn 1
S(t) = 15t³/3+8t+3
S(t) = 5t³+8t+3
b) For the distance traveled by the object between t equals 1 and t equals 2 seconds
When t = 1s
S(1) = 5(1)³/3+8(1)+3
S(1) = 5/3+8+3
S(1) = 5+24+9/3
S(1) = 38/3 m
When t = 2secs
S(2) = 5(2)³/3+8(2)+3
S(2) = 40/3+16+3
S(2) = (40+48+9)/3
S(2) = 97/3
Distance travelled between this times will be 97/3-38/3
= 59/3
= 19.7m approximately
3.7.6. A satellite is in a circular orbit around the Earth. If it is at an altitude equal to twice the radius of the Earth, 2RE, how does its speed v relate to the Earth's radius RE, and the magnitude g of the acceleration due to gravity on the Earth's surface?a)b)c)d)e)
Answer:
Explanation:
For acceleration due to gravity
g = GM / R² ---------------------------- ( 1 )
At height 2R , radius of orbit is 3R .
If speed of satellite be v
m v² / 3R = GMm / (3R)² ( centripetal force = gravitational force )
v² = GM / 3R -------------------------------------- ( 2 )
Dividing (1) and (2)
v² / g = R / 3
v² = gR / 3
v = √( gR / 3 )
Describe what would you do different in your daily routine to eat healthier.....please help me noooow
Answer:
Drink water before every meal. Guzzling a glass of water before every meal can help you lose weight, improve your skin, and feel more energized. ...
Make one of your meals healthier each day. Forget fad diets and strict healthy-eating rules. ...
Walk more. ...
Add vegetables to everything. ...
Get a good night's sleep.
Explanation:
3. Jack and Zack are standing on a crate at rest on a frictionless horizontal surface. Jack has a mass of 70 kg, Zack has a mass of 45 kg, and the crate has a mass of 15 kg. In what follows, we will see that both jump of the crate. You may assume that they push themselves off with a speed of 4 m/s relative to the crate and in a direction that is essentially horizontal.a) what is the final speed of the crate if Jack and Zack jump simultaneously, in the same direction
Answer:
22.4 m/s
Explanation:
To find the speed of the crate you take into account the momentum conservation. The momentum before Jack and Zack jump must be equal to the momentum after. Then you have:
[tex]p_b=p_a\\\\[/tex]
Furthermore, you take into account that the momentum before, when crater, Jack and Zack are at rest, the momentum is zero. Hence:
[tex]p_b=(m_Z+m_J+m_c)v=0\\\\m_Z=45kg\\\\m_J=70kg\\\\m_c=15kg\\\\p_a=m_Zv_Z+m_Jv_J+m_cv_c[/tex]
but you know what are the velocities of Zack and Jack after they jump.
[tex]p_a=p_b=0\\\\-v_c=\frac{m_Zv_Z+m_Jv_J}{m_c}\\\\-v_c=\frac{(70kg)(4m/s)+(45kg)(4m/s)}{15kg}=-22.4\frac{m}{s}[/tex]
the minus sign means that the velocity of the crater is in an opposite direction of the spee of Jack and Zack.
hence, the speed of the crater is 22/4m/s
The final speed of the crate when Jack and Zack jump off simultaneously in the same direction is calculated to be approximately 3.54 m/s. The negative sign indicates opposite direction to the jump.
To solve for the final speed of the crate, we will use the principle of conservation of momentum. Since the system is initially at rest, the total initial momentum is zero.
Let's denote the mass of Jack as mJ, the mass of Zack as mZ, and the mass of the crate as mC. Also, let the velocity of Jack, Zack, and the crate relative to the ground be VJ, VZ, and VC respectively.
[tex]mJ = 70 kg[/tex]
[tex]mZ = 45 kg[/tex]
[tex]mC = 15 kg[/tex]
Since Jack and Zack both push off with a speed of 4 m/s relative to the crate:
VJ = VC + 4 m/s
VZ = VC + 4 m/s
Using conservation of momentum:
Initial momentum = Final momentum
[tex]0 = mJ(VC + 4) + mZ(VC + 4) + mCVC[/tex]
[tex]0 = 70(VC + 4) + 45(VC + 4) + 15VC[/tex]
Combine and solve for VC:
[tex]0 = 70VC + 280 + 45VC + 180 + 15VC[/tex]
[tex]0 = 130VC + 460[/tex]
[tex]-460 = 130VC[/tex]
[tex]VC = -460 / 130[/tex]
[tex]VC = -3.54 m/s[/tex]
The negative sign indicates the crate moves in the opposite direction to their jump. Thus, the final speed of the crate is 3.54 m/s.
Three long wires all lie in an xy plane parallel to the x axis. They intersect the y axis at the origin, y = d, and y = 2d, where d = 10 cm. The two outer wires each carry a current of 6.8 A in the positive x direction.
1. What is the magnitude of the force on a 2.8 m section of either of the outer wires if the current in the center wire is 3.3 A.
(a) in the positive x direction and (b) in the negative x direction?
Answer:
[tex]F = 2.55*10^{-4} \ N[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]
Explanation:
The magnetic field due to the first wire can be written as:
[tex]B_1 = \frac{\mu _o I_1}{2 \pi (2 d)}[/tex]
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
The magnetic field due to the second wire is as follows:
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d}[/tex]
The net magnetic field B = [tex]B_1 +B_2[/tex]
[tex]B = \frac{\mu _o I_1}{4 \pi d} +\frac{\mu _o I_2}{2 \pi d} \\ \\ B = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Also; the magnetic force on wire segment l = 2.8 m
[tex]F = I_1lB = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Replacing all the values ; we have :
[tex]F = \frac{ 4 \pi * 10^{-7}*6.8*2.8}{2 \pi * 10*10^{-2}}(\frac{6.8}{2}+ 3.3)[/tex]
[tex]F = 2.55*10^{-4} \ N[/tex]
b)
Magnetic field due to the first wire is :
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
[tex]B_1 = \frac{4.0*10^{-7}*6.8}{4 \pi *10*10^{-2}}[/tex]
[tex]B_1 = 6.8*10^{-6} \ T[/tex]
Magnetic field due to the second wire is :
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d} \\ \\ B_2 = \frac{4*10^{-7}*3.3}{2 \pi *10*10^{-2}}[/tex]
[tex]B_2 = 6.6*10^{-6}T[/tex]
[tex]B_1>B_2[/tex]
Net magnetic field B = [tex]B_1 - B_2[/tex]
B = [tex](6.8*10^{-6} - 6.6*10^{-6})T[/tex]
B = [tex]2*10^{-7} \ T[/tex]
Magnetic force F = [tex]I_1lB[/tex]
F = [tex]6.6*2.8 *2*10^{-7}[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]
a solenoid has an internal magnetic field of 0.28T. What is the energy density of the field?
Answer:
Energy density, [tex]E=3.11\times 10^4\ J[/tex]
Explanation:
It is given that,
Magnetic field in a solenoid is 0.28 T
We need to find the energy density of the field. The energy density of the magnetic field is given by below formula as follows :
[tex]E=\dfrac{B^2}{2\mu_o}\\\\E=\dfrac{(0.28)^2}{2\times 4\pi \times 10^{-7}}\\\\E=3.11\times 10^4\ J[/tex]
So, the energy density of this field is [tex]3.11\times 10^4\ J[/tex].
What should Jaime do to increase the number of energy storage molecules that producers can make? You may choose more than one answer.
Decrease the amount of carbon dioxide in the ecosystem.
Increase the amount of carbon dioxide in the ecosystem.
Decrease the amount of sunlight in the ecosystem.
Increase the amount of sunlight in the ecosystem.
Answer:
Decrease the amount of carbon dioxide in the ecosystem.
Explanation:
becaus egthy
To increase the number of energy storage molecules that producers can make, Jaime should:
1. Increase the amount of sunlight in the ecosystem: Sunlight is a vital source of energy for photosynthesis, the process by which producers convert carbon dioxide and water into energy-rich molecules such as glucose. More sunlight can potentially increase the rate of photosynthesis, leading to more energy storage molecules being produced.
2. Decrease the amount of carbon dioxide in the ecosystem: While carbon dioxide is essential for photosynthesis, excessive levels can inhibit the process. By decreasing the amount of carbon dioxide in the ecosystem to an optimal level, Jaime can ensure that producers can efficiently utilize available resources to produce energy storage molecules.
So, both increasing sunlight and decreasing carbon dioxide can contribute to increasing the number of energy storage molecules that producers can make.
An infinitely long wire carrying a current of 4.5 A is bent at the origin and extends in both the y-direction and x-direction. (In other words, the wire makes a 90 degree angle with the x axis and y axis and has a bend at the origin) Find the magnetic field at the point x = 3 cm and y = 2 cm.
Answer:
6.444 × [tex]10^{-5}[/tex] T
Explanation:
Find the given attachments
The magnitude of the magnetic field at the point [tex](x = 3 \text{ cm}, y = 2 \text{ cm})[/tex] is approximately [tex]\( 1.08 \times 10^{-5} \text{ T} \)[/tex].
The magnitude of the magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long straight wire carrying a current [tex]\( I \)[/tex] is given by:
[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]
where [tex]\( \mu_0 \)[/tex] is the permeability of free space [tex]\( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)[/tex].
For the horizontal wire, the distance from the wire to the point is [tex]\( y = 2 \text{ cm} = 0.02 \text{ m} \)[/tex]. Thus, the magnetic field due to the horizontal wire is:
[tex]\[ B_x = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})} \][/tex]
Next, consider the vertical section of the wire (along the y-axis). Similarly, the magnetic field due to this section at a perpendicular distance [tex]\( r \)[/tex] is given by the same formula. Here, the distance from the wire to the point is [tex]\( x = 3 \text{ cm} = 0.03 \text{ m} \)[/tex].
The magnetic field due to the vertical wire is:
[tex]\[ B_y = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})} \][/tex]
The total magnetic field [tex]\( B \)[/tex] at the point is the vector sum of [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex]. We can use Pythagoras' theorem to find the magnitude of the resultant magnetic field:
[tex]\[ B = \sqrt{B_x^2 + B_y^2} \][/tex]
Substituting the expressions for [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex] we get:
[tex]\[ B = \sqrt{\left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
Plugging in the value of [tex]\( \mu_0 \)[/tex] and simplifying, we find:
[tex]\[ B = \sqrt{\left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.02 \text{ m}}\right)^2 + \left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.03 \text{ m}}\right)^2} \][/tex]
[tex]\[ B = \sqrt{\left(9 \times 10^{-6} \text{ T}\right)^2 + \left(6 \times 10^{-6} \text{ T}\right)^2} \][/tex]
[tex]\[ B = \sqrt{81 \times 10^{-12} \text{ T}^2 + 36 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B = \sqrt{117 \times 10^{-12} \text{ T}^2} \][/tex]
[tex]\[ B \approx 1.08 \times 10^{-5} \text{ T} \][/tex]