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.
An electron in the first energy level of the electron cloud has an electron in the third energy level
Answer:
a lower energy than
Explanation:
sorry im a month late but is lower energy than
Ratio of the speed of light in a vacuum to the speed of light in a medium Rule for how light is refracted at the boundary between two materials Process that occurs when the angle of incidence is greater than the critical angle
Answer:
TOTAL INTERNAL REFLECTION
Explanation:
Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.
For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.
Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION
When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected
Answer:
The frequency does not change, but the wavelength does
Explanation:
Here are the options
A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.
B. The frequency does change, but the wavelength remains unchanged.
C. Both the frequency and wavelength change.
D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.
E. The frequency does not change, but its wavelength does.
When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.
[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]
where [tex]\lambda_o[/tex] indicates wavelength in vacuum
[tex]\lambda_m[/tex] indicates wavelength in medium
n indicates refractive index
v indicates velocity of light wave
c indicates velocity of light
And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.
Therefore the correct option is E
A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.
Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.
Answer:
v₀(1 + B²L²t/mR)
Explanation:
We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Since F = ma where a = acceleration of loop and m = mass of loop
a = F/m = B²L²v₀/mR
Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0
Substituting the value of a and u into v, we have
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)
The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).
Calculation of the loop velocity:As we know that
Force on the loop
F = BIL
here
B = magnetic field strength,
I = current
and L = length of side of loop.
Now
the current in the loop I = ε/R
where
ε = induced e.m.f in the loop = BLv₀
where v₀ = velocity of loop
and r = resistance of loop
So,
F = BIL = B(BLv₀)L/R = B²L²v₀/R
Also, F = ma where a = acceleration of loop and m = mass of loop
Now
a = F/m = B²L²v₀/mR
We have to use
v = u + at
where
u = initial velocity of loop = v₀,
t = time after t = 0
and v = velocity of loop after time t = 0
So, it be like
v = v₀ + B²L²v₀t/mR
= v₀(1 + B²L²t/mR)
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A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
1) What kind of charged particle was transferred between the rod and the sphere, and in which direction?
A) electrons transferred from rod to sphere.
B) electrons transferred from sphere to rod.
C) protons transferred from rod to sphere.
D) protons transferred from sphere to rod.
2) How many charged particles were transferred?
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
1. Electrons transferred from sphere to rod.
Option B is correct.
2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.
Given that initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let us consider that the charge absorbed by the plastic rod is q
[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Therefore, electrons transferred from sphere to rod
The charge on one electron is, 1.602 x 10⁻¹⁹ C .
Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]
Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.
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6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?
Answer:
[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]
Given:
Mass of oxygen (m) = 32.0 g = 0.032 kg
Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³
Speed of sound in oxygen (v) = 317 m/s
To Find:
Bulk modulus of oxygen
Explanation:
[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]
[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]
[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]
[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]
[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]
[tex]\sf \implies B = 100489 \times 1.428[/tex]
[tex]\sf \implies B = 143498.292 \: Pa[/tex]
[tex]\sf \implies B \approx 143.5 \: kPa[/tex]
Suppose you are playing hockey on a new-age ice surface for which there is no friction between the ice and the hockey puck. You wind up and hit the puck as hard as you can. After the puck loses contact with your stick, the puck will
Answer:
Not slow down or speed up.
Explanation:
Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.
Consider 1 mol an ideal gas at 28∘ C and 1.06 atm pressure. To get some idea how close these molecules are to each other, on the average, imagine them to be uniformly spaced, with each molecule at the center of a small cube.
A) What is the length of an edge of each cube if adjacent cubes touch but do not overlap?
B) How does this distance compare with the diameter of a typical molecule? The diameter of a typical molecule is about 10-10 m. (in l/dmolecule)
C) How does their separation compare with the spacing of atoms in solids, which typically are about 0.3 nm apart? (in l/lsolid)
Answer:
A) Length of an edge = 3.38 × 10^(-9) m
B) 34 times the diameter of a molecule.
C) 11 times the atomic spacing in solids.
Explanation:
A) We will use Avogadro's hypothesis to solve this. It states that 1 mole of gas occupies 22.4 L at STP.
We want to find the volume occupied by 1 mole of gas at 1.06 atm pressure and temperature of 28 °C (= 301 K).
Thus, by the ideal gas equation, we have;
V_mole = (1 × 22.4/273) × (301/1.06) = 23.3 L = 0.0233 m³
Now, since from avogadros number, 1 mole of gas contains 6.02 x 10^(23) molecules, then volume occupied by a molecule is given by;
V_molecule = 0.0233/(6.02 × 10^(23)) m³ = 3.87 x 10^(-26) m³
Thus, length of an edge of the cube = ∛(3.87 × 10^(-26)) = 3.38 × 10^(-9) m
B) We are told that The diameter of a typical molecule is about 10^(-10) m.
Thus, the distance is about;
(3.38 × 10^(-9))/(10^(-10)) ≈ 34 times the diameter of a molecule.
C) We are told that the spacing of atoms is typically are about 0.3 nm apart
Thus;
The separation will be about;
(3.38 × 10^(-9))/(0.3 × 10^(-9)) ≈ 11 times the atomic spacing in solids.
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Answer
Work done is 57.9KJ
Explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.
Answer:
b. approaches infinity
Explanation:
Because Capacitive reactance is given as Xc = 1/ωC
So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.
Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC
1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?
Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.
Explanation:
Answer:
Light amplification by stimulated emission of radiation
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 concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.
Answer:
D.
Inverted and reduced
If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.
A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
Hence option B is correct.
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The primary difference between a barometer and a manometer is
A. a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
B. a barometer uses mercury, while a manometer can use any liquid. a barometer is used to measure atmospheric pressure, and a manometer is used to measure absolute pressure.
C a barometer reads in mm, while a manometer reads in Pa.
D a barometer can measure either positivee or negative pressure, while a manometer only
E positive pressure. measures
Answer:
a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
Explanation:
A barometer measures air pressure at any locality with sea level as the reference.
However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure
The algebraic sum of these two values gives the absolute pressure.
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.
Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
Which does not account for the fact that fish can survive the winter in ponds in temperate climate zones? 1. the density of ice versus that of water 2. the unique properties of water 3. the intermolecular bonding of water 4. the tendency for water to freeze from the bottom up
Answer:
3. the intermolecular bonding of water
Explanation:
Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.
Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.
Answer:
D. The tendency for water to freeze from the bottom up.
Explanation:
The orbital motion of Earth around the Sun leads to an observable parallax effect on the nearest stars. For each star listed, calculate the distance in parsecs before converting that distance to astronomical units. A. Sirius (0.38") B. Alpha Centauri A (0.75") C. Procyon (0.28") D. Wolf 359 (0.42") E. Epsilon Eridani (0.31") D(pc) = 1/parallax(arcsecs), D(a.u.) = D(pc) * 206265 (arcsecs per radian)
Answer:
Following are the answer to this question:
Explanation:
Formula:
[tex]D(PC) =\frac{1}{parallax}\\\\D(av)=D(PC) \times 20.626\ J[/tex]
Calculating point A:
when the value is [tex]0.38[/tex]
[tex]\to 0.38 \toD(PC)= \frac{1}{0.38}\\\\[/tex]
[tex]=2.632[/tex]
[tex]\to D(a.v) = \frac{1}{0.38} \times 206265\\[/tex]
[tex]=542,802.6[/tex]
Calculating point B:
when the value is [tex]0.75[/tex]
[tex]\to D(PC)=\frac{1}{0.75}[/tex]
[tex]=1.33[/tex]
[tex]\to D(a.v) = \frac{1}{0.75} \times 206265\\[/tex]
[tex]=275,020[/tex]
Calculating point C:
when the value is [tex]0.28[/tex]
[tex]\to D(PC)=\frac{1}{0.28}[/tex]
[tex]=3.571[/tex]
[tex]\to D(a.v) = \frac{1}{0.28} \times 206265\\[/tex]
[tex]=736660.7[/tex]
Calculating point D:
when the value is [tex]0.42[/tex]
[tex]\to D(PC)=\frac{1}{0.42}[/tex]
[tex]=2.38[/tex]
[tex]\to D(a.v) = \frac{1}{0.42} \times 206265\\[/tex]
[tex]=490910.7[/tex]
Calculating point E:
when the value is [tex]0.31[/tex]
[tex]\to D(PC)=\frac{1}{0.31}[/tex]
[tex]=3.226[/tex]
[tex]\to D(a.v) = \frac{1}{0.31} \times 206265\\[/tex]
[tex]=665370.97[/tex]
A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston
Answer:
The cross-sectional area of the larger piston is 392 cm²
Explanation:
Given;
output mass of the piston, m₀ = 2000 kg
input force of the piston, F₁ = 500 N
input area of the piston, A₁ = 10 cm² = 0.001 m²
The output force is given by;
F₀ = m₀g
F₀ = 2000 x 9.8
F₀ = 19600 N
The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;
[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]
Therefore, the cross-sectional area of the larger piston is 392 cm²
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
Explain how surface waves can have characteristics of both longitudinal waves and transverse waves. Please use 3 content related sentences
Answer: Search Results
Featured snippet from the web
Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.
Explanation:
Surface waves can exhibit characteristics of both longitudinal waves and transverse waves.
Surface waves are a type of mechanical wave that propagate along the interface between two different mediums, such as the ground and air or the surface of water. These waves combine properties of both longitudinal and transverse waves
Similar to longitudinal waves, surface waves involve particles oscillating in the same direction as the wave propagation. This creates compressions and rarefactions, leading to variations in density or pressure. These compressions and rarefactions are characteristic of longitudinal waves.
However, surface waves also exhibit transverse motion. As the wave propagates along the surface, particles move in a perpendicular direction to the wave's motion. This transverse motion causes particles to displace vertically or horizontally, similar to transverse waves.
By combining both longitudinal and transverse characteristics, surface waves possess a complex motion that allows them to travel along the surface while simultaneously causing particles to oscillate both parallel and perpendicular to the direction of wave propagation.
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help... Please help!!!!!!!!!!!
Answer:
a) 6.8--5.10 thats equal 11.9
b) m=ris/run +10 equal 0.06/8 =7.5*10^-3
A rod on a compressed spring exerts 12 N of force on a 0.05-kg steel ball. The
rod pushes the ball 0.03 m. How much work does the spring do on the ball?
A) 36
B) 36 N
C) 60 N
D)1.00
Answer:
Work = 0.36N
Explanation:
Given
Force = 12N
Distance = 0.03m
Weight = 0.05kg
Required
Determine the work done
Workdone is calculated as thus;
Work = Force * Distance
Substitute 12N for Force and 0.03m for Distance
Work = 12N * 0.03m
Work = 0.36Nm
Using proper S.I units
Work = 0.36N
Hence, work done by the spring on the ball is 0.36N
A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field
Answer:
The amplitude of the oscillating electric field is 1316.96 N/C
Explanation:
Given;
frequency of the wave, f = 2.4 Hz
intensity of the wave, I = 2300 W/m²
Amplitude of oscillating magnetic field is given by;
[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is intensity of wave
c is speed of light = 3 x 10⁸ m/s
[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]
The amplitude of the oscillating electric field is given by;
E₀ = cB₀
E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶
E₀ = 1316.96 N/C
Therefore, the amplitude of the oscillating electric field is 1316.96 N/C
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Complete Question
A 590-turn solenoid is 12 cm long. The current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?
Answer:
The force is [tex]F = 0.1602 \ N[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 590 \ turns[/tex]
The length of the solenoid is [tex]L = 12 \ cm = 0.12 \ m[/tex]
The current is [tex]I = 36 \ A[/tex]
The diameter is [tex]D = 4.5 \ cm = 0.045 \ m[/tex]
The current carried by the wire is [tex]I = 27 \ A[/tex]
The length of the wire is [tex]l = 2 cm = 0.02 \ m[/tex]
Generally the magnitude of the force on this wire assuming the solenoid's field points due east is mathematically represented as
[tex]F = B * I * l[/tex]
Here B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{L}[/tex]
Here [tex]\mu _o[/tex] is permeability of free space with value [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]
substituting values
[tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]
[tex]B = 0.2225 \ T[/tex]
So
[tex]F = 0.2225 * 36 * 0.02[/tex]
[tex]F = 0.1602 \ N[/tex]
A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?
Answer:
a) 0.988 m/s
b) 199512 Pa
c) 57.52 s
Explanation:
given that
A
A1 v1 = A2 v2
d1² v1 = d2² v2
v2 = [d1/d2]² v1
v2 = (1.2/5.4)² * 20
v2 = 0.049 * 20
v2 = 0.988 m/s
B
P + 1/2 ρ v² = K.
[p2 - p1] = 1/2 ρ [v1² - v2²]
[p2 - p1] = 1/2 * 1000 [20² - 0.988²]
[p2 - p1] = 500 * (400 - 0.976)
[p2 - p1] = 500 * 399
[p2 - p1] = 199512 Pa
C
Flow rate = AV = π [d²/ 4 ] * v
= π [0.012² / 4 ] * 20 = 0.00226 m³ /s
= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s
130 liters = 0.13 m³
t = 0.13/ 0.00226 = 57.52 s
A homeowner purchases insulation for her attic rated at R-15. She wants the attic insulated to R-30. If the insulation she purchased is 10 cm thick, what thickness does she need to use
Answer:
she need to use 20 cm thick
Explanation:
given data
wants the attic insulated = R-30
purchased = 10 cm thick
solution
as per given we can say that
10 cm is for the R 15
but she want for R 30
so
R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]
R 30 thickness = 20 cm
so she need to use 20 cm thick
A man using a 70kg garden roller on a level surface, exerts a force of 200N at 45 degrees to the ground. find the vertical force of the roller on the ground if,
i.he pulls
ii.he pushes the roller
Answer:
i) 545.2 N upwards
ii) 828.2 N downwards
Explanation:
mass of the roller = 70 kg
force exerted = 200 N
angle the force makes with the ground ∅ = 45°
weight of the roller W = mg
where
m is the mass of the roller
g is the acceleration due to gravity = 9.81 m/s^2
weight of the roller = 70 x 9.81 = 686.7 N
The effective vertical force exerted by the man = F sin ∅ = 200 x sin 45°
==> F = 200 x 0.707 = 141.5 N
i) if the man pulls, then the exerted force will be in opposite direction to the weight of the roller vertically upwards
Resultant vertical force = 686.7 N - 141.5 N = 545.2 N upwards
ii) if he pushes, then the exerted force will be in the direction of the weight vertically downwards
Resultant vertical force = 686.7 N + 141.5 N = 828.2 N downwards
You are holding on to one end of a long string that is fastened to a rigid steel light pole. After producing a wave pulse that was 5 mm high and 4 em wide, you want to produce a pulse that is 4 cm wide but 7 mm high. You must move your hand up and down once,
a. a smaller distance up, but take a shorter time.
b. the same distance up as before, but take a shorter time.
c. a greater distance up, but take a longer time.
d. the same distance up as before, but take a longer time.
e. a greater distance up, but take the same time.
Answer:
It will take. the same distance up as before, but take a longer time
"Can we consider light wave as a single frequency wave? Either Yes or No, explain the reason of your answer. "
Answer:
Well, yes.
We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.
Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.
An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.
Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.
