Answer:
In parallel
Explanation:
Ctotal = C1 + C2 + ... + Cn
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
Find out more information about magnetic field here:
https://brainly.com/question/26257705
A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish
Answer:
2,66
Explanation:
The refractive index= real depth/ apparent depth
real depth = refractive index * apparent depth
Let's assume index for water is 1.33
real depth = 2*1,33 = 2,66
What is the difference between matter and energy
Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
A thick wire with a radius of 4.0 mm carries a uniform electric current of 1.0 A, distributed uniformly over its cross-section. At what distance from the axis of the wire, and greater than the radius of the wire, is the magnetic field strength equal to that at a distance 2.0 mm from the axis. distance
Answer:
8 mm
Explanation:
From the information given:
The Ampere circuital law can be used to estimate the magnetic field strength at two points when the distance is less than the radius and when the distance is greater than the radius.
when the distance is less than the radius ; we have:
[tex]B_1 = \dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2}[/tex]
when the distance is greater than the radius; we have:
[tex]B_2 = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
Equating both equations together ; we have :
[tex]\dfrac{ \mu_o \times I \times r}{2 \times \pi \times \ d^2} = \dfrac{\mu_o \ \times I}{2 \ \times \pi \times R}[/tex]
[tex]\dfrac{1}{R}= \dfrac{r}{d^2}[/tex]
[tex]R= \dfrac{d^2}{r}[/tex]
where; d = radius of the wire and r = distance;
[tex]R =\dfrac{4^2}{2}[/tex]
[tex]R =\dfrac{16}{2}[/tex]
R = 8 mm
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density
Answer:
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
Explanation:
given data
frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz
we consider here intensity I = 0.2 W/m²
solution
we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s
and take energy density is E
so here
E × C = I
E = [tex]\frac{I}{C}[/tex] ................1
here C = 3 × [tex]10^{8}[/tex] m/s
so photon density is
photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex] ...............2
photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]
photon density = 1.0 × [tex]10^{16}[/tex] photon/m³
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
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 rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 10.8 N . What is the smallest density of a liquid in which the rock will float?
Answer:
The density is [tex]\rho_z = 2544 \ kg /m^3[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 1.80 \ kg[/tex]
The tension on the string is [tex]T = 10.8 \ N[/tex]
Generally the weight of the rock is
[tex]W = m * g[/tex]
=> [tex]W = 1.80 * 9.8[/tex]
=> [tex]W = 17.64 \ N[/tex]
Now the upward force(buoyant force) acting on the rock is mathematically evaluated as
[tex]F_f = W - T[/tex]
substituting values
[tex]F_f = 17.64 - 10.8[/tex]
[tex]F_f = 6.84 \ N[/tex]
This buoyant force is mathematically represented as
[tex]F_f = \rho * g * V[/tex]
Here [tex]\rho[/tex] is the density of water and it value is [tex]\rho = 1000\ kg/m^3[/tex]
So
[tex]V = \frac{F_f}{ \rho * g }[/tex]
[tex]V = \frac{6.84}{ 1000 * 9.8 }[/tex]
[tex]V = 0.000698 \ m^3[/tex]
Now for this rock to flow the upward force (buoyant force) must be equal to the length
[tex]F_f = W[/tex]
[tex]\rho_z * g * V = W[/tex]
Here z is smallest density of a liquid in which the rock will float
=> [tex]\rho_z = \frac{W}{ g * V}[/tex]
=> [tex]\rho_z = \frac{17.64}{ 0.000698 * 9.8}[/tex]
=> [tex]\rho_z = 2544 \ kg /m^3[/tex]
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.
For this "type" of motion, displacement (Δx) can be determined by:
[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]
[tex]v_{i}[/tex] is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]
[tex]\Delta x = 30 - 2.36[/tex]
[tex]\Delta x[/tex] = - 42
Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.
A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it discharges a 15 μF capacitor through paddles placed on the skin, causing charge to flow through the heart. Assume that the capacitor is originally charged with 5.0 kV .Part AWhat is the charge initially stored on the capacitor?3×10−9 C7.5×104 C7.5×10−2 C7.5×10−5 CPart BWhat is the energy stored on the capacitor?What is the energy stored on the capacitor?1.9×108 J380 J190 J1.9×10−4 JPart CIf the resistance between the two paddles is 100 Ω when the paddles are placed on the skin of the patient, how much current ideally flows through the patient when the capacitor starts to discharge?5×105 A50 A2×10−2 A5×10−2 APart DIf a defibrillator passes 17 A of current through a person in 90 μs . During this time, how much charge moves through the patient?If a defibrillator passes 17 {\rm A} of current through a person in 90 {\rm \mu s} . During this time, how much charge moves through the patient?190 mC1.5 C1.5 mC17 C
Answer:
a) q = 7.5 10⁻² C , b) 190 J , c) I₀ = 50 A , d) 1.5 mC
Explanation:
The expression for capacitance is
C = q / DV
q = C DV
let's reduce the magnitudes to the SI system
ΔV = 5 kV = 5000 V
C = 15 μF = 15 10⁻⁶ F
t = 90 μs = 90 10⁻⁶ s
q = 15 10⁻⁶ 5000
q = 7.5 10⁻² C
b) the energy in a capacitor is
U = ½ C ΔV²
U = ½ 15 10⁻⁶ 5000²
U = 1,875 10² J
answer 190 J
c) At the moment the discharge begins, all the current is available and it decreases with time,
whereby
V = I R
in the first instant I = Io
I₀ = V / R
I₀ = 5000/100
I₀ = 50 A
but this is for a very short time
answer 50 A
d) The definition of current is
i = dq / dt
in this case they give us the total current and the total time, so we can find the total charge
i = q / t
q = i t
q = 17 90 10⁻⁶
q = 1.53 10⁻³ C
answer is 1.5 mC
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]
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
A pair of narrow, parallel slits separated by 0.230 mm is illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.50 m away from the plane of the parallel slits.
A) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
B) Calculate the distance between the first and second dark bands in the interference pattern.
Answer:
A) y = 3.56 mm
B) y = 3.56 mm
Explanation:
A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:
[tex] y = \frac{m\lambda L}{d} [/tex]
Where:
λ: is the wavelength = 546.1 nm
m: is first bright region = 1
L: is the distance between the screen and the plane of the parallel slits = 1.50 m
d: is the separation between the slits = 0.230 mm
[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
B) The distance between the first and second dark bands is:
[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]
Where:
[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]
[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]
I hope it helps you!
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Learn more about the average density here:
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Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
IMPORTANT ANSWER ALL 3 PLEASE!
Answer:
4. Liters
5. Celsius
6. Grams
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.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°
Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?
Explanation:
Given that,
Linear speed of both disks is 5 m/s
Mass of disk 1 is 10 kg
Radius of disk 1 is 35 cm or 0.35 m
Mass of disk 2 is 3 kg
Radius of disk 2 is 7 cm or 0.07 m
(a) The angular velocity of disk 1 is :
[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]
(b) The angular velocity of disk 2 is :
[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]
(c) The moment of inertia for the two disk system is given by :
[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]
Hence, this is the required solution.
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
Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answer:
They both have the same acceleration
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 simple arrangement by means of which e.m.f,s. are compared is known
Answer:
A simple arrangement by means of which e.m.f,s. are compared is known as?
(a)Voltmeter
(b)Potentiometer
(c)Ammeter
(d)None of the above
Explanation:
Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
[tex]m1U1+ m2U2= m1V1+m2V2[/tex]
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
[tex]m1U1+ m2U2= V(m1+m2)[/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]
A 23 cm tall object is placed in front of a concave mirror with a radius of 37 cm. The distance of the object to the mirror is 86 cm. Calculate the focal length of the mirror.
Answer:
18.5 cm
Explanation:
From;
1/u + 1/v = 1/f
Where;
u= object distance = 86cm
image height = 23 cm
Radius of curvature = 37 cm
The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.
Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm
Therefore, the focal length of the mirror is 18.5 cm
Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the universe. The universe is very big and very old. The universe consists of matter.
Answer:
Explanation:
That Universe Consists of Matter
A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached
Answer:
The displacement is [tex]A_r = 6.071 \ cm[/tex]
Explanation:
From the question we are told that
The initial displacement is [tex]A_o = 10 \ cm[/tex]
The displacement at the end of first oscillation is [tex]A_d = 9.05 \ cm[/tex]
Generally the damping constant of this damped oscillator is mathematically represented as
[tex]\eta = \frac{A_d}{A_o}[/tex]
substituting values
[tex]\eta = \frac{9.05}{10}[/tex]
[tex]\eta = 0.905[/tex]
The displacement after 4 more oscillation is mathematically represented as
[tex]A_r = \eta^4 * A_d[/tex]
substituting values
[tex]A_r = (0.905)^4 * (9.05)[/tex]
[tex]A_r = 6.071 \ cm[/tex]
Answer:
Displacement reached is 6.0708 cm
Explanation:
Formula for damping Constant "C"
[tex]C^n=\frac{A_2}{A_1}[/tex] where n=1,2,3,........n
Where:
[tex]A_2[/tex] is the displacement after first oscillation
[tex]A_1\\[/tex] is the initial Displacement
[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]
In our case, n=1.
[tex]C=\frac{9.05}{10}\\C=0.905[/tex]
After 4 more oscillation, n=4:
[tex]C^4=\frac{A_6}{A_2}[/tex]
Where:
[tex]A_6[/tex] is the final Displacement after 4 more oscillations.
[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]
Displacement reached is 6.0708 cm
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
