The correct answer is D. An average city
Explanation:
A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.
A 0.100-kg metal rod carrying a current of 15.0 A glides on two horizontal rails 0.550 m apart and 2.0 m long,
(a) If the coefficient of kinetic friction between the rod and rails is 0.120, what vertical magnetic field is required to keep the rod moving at a constant speed?
(b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).
Answer:
The speed of the rod is 2.169 m/s.
Explanation:
Given that,
Mass = 0.100 kg
Current = 15.0 A
Distance = 2 m
Length = 0.550 m
Kinetic friction = 0.120
(a). We need to calculate the magnetic field
Using relation of frictional force and magnetic force
[tex]F_{f}=F_{B}[/tex]
[tex]\mu mg=Bli[/tex]
[tex]B=\dfrac{\mu mg}{li}[/tex]
Where, l = length
i = current
m = mass
Put the value into the formula
[tex]B=\dfrac{0.120\times0.1\times9.8}{0.550\times15.0}[/tex]
[tex]B=0.01425\ T[/tex]
[tex]B=1.425\times10^{-2}\ T[/tex]
(b). If the friction between the rod and rail is reduced zero.
So, [tex]f_{f}=0[/tex]
We need to calculate the acceleration
Using formula of force
[tex]F_{net}=f_{f}+F_{B}[/tex]
[tex]F_{net}=0+Bil[/tex]
[tex]ma=Bil[/tex]
[tex]a=\dfrac{Bil}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.425\times10^{-2}\times15\times0.55}{0.1}[/tex]
[tex]a=1.176\ m/s^2[/tex]
We need to calculate the speed of the rod
Using equation of motion
[tex]v^2=u^2+2as[/tex]
Put the value into the formula
[tex]v^2=0+2\times1.176\times2[/tex]
[tex]v^2=\sqrt{4.704}\ m/s[/tex]
[tex]v=2.169\ m/s[/tex]
Hence, The speed of the rod is 2.169 m/s.
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
What is the direction of the net gravitational force on the mass at the origin due to the other two masses?
Answer:
genus yds it's the
Explanation:
xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj
A skull believed to belong to an ancient human being has a carbon-14 decay rate of 5.4 disintegrations per minute per gram of carbon (5.4 dis/min*gC). If living organisms have a decay rate of 15.3 dis/min*gC, how old is this skull
Answer:
9.43*10^3 year
Explanation:
For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation
To start with, we use the formula
t(half) = In 2/k,
if we make k the subject of formula, we have
k = in 2/t(half), now we substitute for the values
k = in 2 / 5730
k = 1.21*10^-4 yr^-1
In(A/A•) = -kt, on rearranging, we find out that
t = -1/k * In(A/A•)
The next step is to substitite the values for each into the equation, giving us
t = -1/1.21*10^-4 * In(5.4/15.3)
t = -1/1.21*10^-4 * -1.1041
t = 0.943*10^4 year
The Milky Way has a diameter (proper length) of about 1.2×105 light-years. According to an astronaut, how many years would it take to cross the Milky Way if the speed of the spacecraft is 0.890 c?
Answer:
t = 134834.31 years
Explanation:
First we find the speed of the ship:
v = 0.890 c
where,
v = speed of the ship = ?
c = speed of light = 3 x 10⁸ m/s
Therefore, using the values, we get:
v = (0.89)(3 x 10⁸ m/s)
v = 2.67 x 10⁸ m/s
Now, we find the distance in meters:
Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)
s = 11.35 x 10²⁰ m
Now, for the time we use the following equation:
s = vt
t = s/v
t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)
t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)
t = 134834.31 years
A 1.2-m length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x= 5.0m on x-axis.
a. 1.6 nt in the negative z direction
b. 1.6 nt in the positive z direction
c. 2.4 T in the positive z direction
d. 2.4 nt in the negative z direction
e. None of the above
Answer:
None of the above
Explanation:
The formula of the magnetic field of a point next to a wire with current is:
B = 2×10^(-7) × ( I /d)
I is the intensity of the current.
d is the distance between the wire and the point.
● B = 2*10^(-7) × (20/5) = 8 ×10^(-7) T
Structures on a bird feather act like a diffraction grating having 8500 lines per centimeter. What is the angle of the first-order maximum for 577 nm light shone through a feather?
Answer:
29.5°
Explanation:
To find the distance d
d = 1E10^-2/8500lines
= 1.17x 10-6m
But wavelength in first order maximum is 577nm
and M = 1
So
dsin theta= m. Wavelength
Theta= sin^-1 (m wavelength/d)
= Sin^-1 ( 1* 577 x10^-8m)/1.17*10^-6
= 493*10^-3= sin^-1 0.493
Theta = 29.5°
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F⃗ to it. As a result, the block moves along the x-axis such that its position as a function of time is given by x(t)=αt2+βt3, where α = 0.210 m/s2 and β = 2.04×10−2 m/s3 .
A. Calculate the velocity of the object at time t = 4.50 s .
B. Calculate the magnitude of F⃗ at time t = 4.50 s .
Express your answer to three significant figures.
C. Calculate the work done by the force F⃗ during the first time interval of 4.50 s of the motion.
Express your answer to three significant figures.
Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
One solenoid is centered inside another. The outer one has a length of 54.0 cm and contains 6750 coils, while the coaxial inner solenoid is 4.00 cm long and 0.170 cm in diameter and contains 21.0 coils. The current in the outer solenoid is changing at 35.0 A/s .What is the mutual inductance of the solenoids?Find the emf induced in the inner solenoid.
Answer:
M₁₂ = 1.01 10⁻⁴ H , Fem = 3.54 10⁻³ V
Explanation:
The mutual inductance between two systems is
M₁₂ = N₂ Ф₁₂ / I₁
where N₂ is the number of turns of the inner solenoid N₂ = 21.0, i₁ the current that flows through the outer solenoid I₁ = 35.0 A / s and fi is the flux of the field of coil1 that passes through coil 2
the magnetic field of the coil1 is
B = μ₀ n I₁ = μ₀ N₁/l I₁
the flow is
Φ = B A₂
the area of the second coil is
A₂ = π d₂ / 4
Φ = μ₀ N₁ I₁ / L π d² / 4
we substitute in the first expression
M₁₂ = N₂ μ₀ N₁ / L π d² / 4
M₁₂ = μ₀ N₁ N₂ π d² / 4L
d = 0.170 cm = 0.00170 m
L = 4.00 cm = 0.00400 m
let's calculate
M₁₂ = 4π 10⁻⁷ 6750 21 π 0.0017²/ (4 0.004)
M₁₂ = π² 0.40966 10⁻⁷ / 0.004
M₁₂ = 1.01 10⁻⁴ H
The electromotive force is
Fem = - M dI₁ / dt
Fem = - 1.01 10⁻⁴ 35.0
Fem = 3.54 10⁻³ V
Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)
d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)
= 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
(A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Explanation:
Given that,
Magnetic field = 0.45 T
The loop's diameter changes from 17.0 cm to 6.0 cm .
Time = 0.53 sec
(A). We need to find the direction of the induced current.
Using Lenz law
If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.
(B). We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{1}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{1}=0.01021\ Wb[/tex]
We need to calculate the magnetic flux
Using formula of flux
[tex]\phi_{2}=BA\cos\theta[/tex]
Put the value into the formula
[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]
[tex]\phi_{2}=0.00127\ Wb[/tex]
We need to calculate the magnitude of the average induced emf
Using formula of emf
[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]
Put the value into t5he formula
[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]
[tex]\epsilon=0.016867\ V[/tex]
[tex]\epsilon=16.87\ mV[/tex]
(C). If the coil resistance is 2.5 Ω.
We need to calculate the induced current
Using formula of current
[tex]I=\dfrac{\epsilon}{R}[/tex]
Put the value into the formula
[tex]I=\dfrac{0.016867}{2.5}[/tex]
[tex]I=0.00675\ A[/tex]
[tex]I=6.75\ mA[/tex]
Hence, (A). The direction of the induced current will be clockwise.
(B). The magnitude of the average induced emf 16.87 mV.
(C). The induced current is 6.75 mA.
Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m
Answer:
0.03/π m/min
Explanation:
See attached file pls
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
Which of the following regions of the electromagnetic spectrum have longer wavelengths than visible light? 1. infrared radiation 2. ultraviolet radiation 3. microwave radiation
Answer:infrared radiation
Explanation:
Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
What is electromagnetic wave?EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.
If we arrange electromagnetic wave with decrease in wavelength, we get:
Radio waves > microwave > Infrared > Visible light > Ultraviolet > X-rays > Gamma radiation.
Hence, Infrared radiation and microwave radiation of the electromagnetic spectrum have longer wavelengths than visible light.
Learn more about electromagnetic wave here:
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Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
Can anyone provide me the answer with explanation?
Answer:
the answer to your question us c honey
Answer:
C
Explanation:
This is so because different materials vary in resistance and conductance of current, heat. Metals are good conductors while none metals like rubber, plastic, glass etc are good insulators or resistors.
Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!
Answer:
Hey there!
This can be a confusing topic, so it's totally fine if you get confused...
First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.
Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.
Let me know if this helps :)
A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf
Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?
Express your answer with the appropriate units.
Answer:
The induced emf is [tex]\epsilon = 0.1041 \ V[/tex]
Explanation:
From the question we are told that
The radius of the circular loop is [tex]r = 9.50 \ cm = 0.095 \ m[/tex]
The intensity of the wave is [tex]I = 0.0215 \ W/m^2[/tex]
The wavelength is [tex]\lambda = 6.90\ m[/tex]
Generally the intensity is mathematically represented as
[tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
B is the magnetic field which can be mathematically represented from the equation as
[tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]
substituting values
[tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]
[tex]B = 1.342 *10^{-8} \ T[/tex]
The area is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.095)^2[/tex]
[tex]A = 0.0284[/tex]
The angular velocity is mathematically represented as
[tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]
substituting values
[tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]
[tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = N * B * A * w * sin (wt )[/tex]
At maximum induced emf [tex]sin (wt) = 1[/tex]
So
[tex]\epsilon = N * B * A * w[/tex]
substituting values
[tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]
[tex]\epsilon = 0.1041 \ V[/tex]
A magnetic field near the floor points down and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counter-clockwise?
a. clockwiseb. counter-clockwise c. no curly E
Answer:
when a magnetic field near the floors points down and is increasing then the electric field curl (a) clockwise.
Explanation:
The magnetic field this is the area that is around a magnet which there is presence of magnetic force. The Moving electric charges can create magnetic fields. we say In physics, that the magnetic field is a field that passes through space and which makes a magnetic force move electric charges.
The Non-coulomb electric field curls ; ( B ) counterclockwise
Non-coulomb electric field also known as induced EMF is the Negative time rate of change of a magnetic flux in a closed loop through the loop. Non-coulomb electric field is expressed as ; Fnc = qEnc
Given that the magnetic field points downwards and the value of the electric field ( ε ) is increasing ( i.e. ε > 0 ) The direction of the non-coulomb electric field will curl in a counter-clockwise direction.
Hence we can conclude that The Non-coulomb electric field curls in a counterclockwise direction.
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"When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is"
Complete Question
When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is [tex]36.0 ^o[/tex] . What are
(a) the index of refraction of the glass and
(b) the Brewster angle?
Answer:
a
[tex]n_r = 1.376[/tex]
b
[tex]i = 54^o[/tex]
Explanation:
From the question we are told that
The angle of refraction is [tex]r = 36.0 ^o[/tex]
Generally according Brewster law
[tex]i + r = 90[/tex]
Here [tex]i[/tex] is the angle of incidence which is also the Brewster angle
So
[tex]i + 36.0 = 90[/tex]
[tex]i = 54^o[/tex]
Now the refractive index is mathematically represented as
[tex]n_r = tan (i)[/tex]
substituting values
[tex]n_r = tan (54)[/tex]
[tex]n_r = 1.376[/tex]
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
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A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have
Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex] (1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex] (2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
A sphere of radius R has charge Q. The electric field strength at distance r > R is Ei.
What is the ratio Ef /Ei of the final to initial electric field strengths if (a) Q is halved, (b) R is halved, and (c) r is halved (but is still > R)? Each part changes only one quantity; the other quantities have their initial values.
Answer:
A. Ef/ Ei = 1/2
B. EF/ Ei = 1
C Ef / Ei = 4
Explanation:
To solve this we apply Coulomb's law which States that
E = Kq / r^2
Where
q = charge r = straight line distance from q to the point in question and
K = Coulomb's constant
Then
Ei = K Q / r^2
So
A) If Q is halved then
Ef = K Q / (2 r^2)
Ef/Ei = 1/2
B) If R is halved, the value of the E-f
at a distance r remains unchanged. So
Ef/Ei = 1
C) if r is now r/2 then
Ef = K Q / (r/2)^2 = K Q / r^2/4 = 4 K Q / r^2
Ef / Ei = 4
If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
