Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 μF capacitor from a camera flash unit retains a voltage of 140 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answer:
93.3x10^-3s
Explanation:
If
Resistance = 1.8 kΩ
Current = 50 mA
Capacitor = 120 μF
Voltage = 140 V
to calculate the discharge current
Applying the formula of discharge current
io=vo/R
io= 140/ 1.8x 10³
= 0.078A
to calculate the time
Applying the formula of current
io= vo/R e-t/RC
50= 140/1800e-t/RC
0.649= e-t/RC
-t/RC= ln( 0.649)
t = 0.432x 120x10^-6x 1800
t=93.3 x 10^-3seconds
Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).
Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answer:
a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².
Explanation:
a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.
τ = Frsinθ
= 185 N × 1.73 m × sin90°
= 320.05 Nm
So, the torque τ = 320.05 Nm
b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m
α = angular acceleration of grindstone
τ = Iα
α = τ/I = τ/(MR²/2) = 2τ/MR²
substituting the values of the variables, we have
α = 2τ/MR²
= 2 × 320.05 Nm/[76 kg × (1.73 m)²]
= 640.1 Nm/227.4604 kgm²
= 2.814 rad/s²
So, the angular acceleration α = 2.814 rad/s²
c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.
So τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm
The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm
Since τ'' = Iα
α' = τ''/I where α' = its new angular acceleration
α' = 2τ/MR²
= 2 × 319.75 Nm/[76 kg × (1.73 m)²]
= 639.5 Nm/227.4604 kgm²
= 2.811 rad/s²
So, the angular acceleration α' = 2.811 rad/s²
Find the focal length of contact lenses that would allow a nearsighted person with a 130 cmcm far point to focus on the stars at night.
Answer:
130cmExplanation:
The lens equation is expressed as;
1/f = 1/u+1/v where;
f is the focal length of the lens
u is the object distance
v is the image distance
Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.
The object distance u = 130cm
Substituting the given parameters in the formula to get the focal length f
[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]
[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]
[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]
Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm
An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
I = 1.23 A
Explanation:
In an RL circuit current passing is described by
I = E / R (1 - [tex]e^{-Rt/L}[/tex])
Let's reduce the magnitudes to the SI system
L = 35 mH = 35 10⁻³ H
t = 5.0 ms = 5.0 10⁻³ s
let's calculate
I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))
I = 1.5 (1- [tex]e^{-1.715}[/tex])
I = 1.23 A
Which was a major effect of Pope Leo III crowning Charlemagne emperor of the Romans ?
Answer:
The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.
You make a telephone call from New York to a friend in London. Estimate how long it will take the electrical signal generated by your voice to reach London, assuming the signal is (a) carried on a telephone cable under the Atlantic Ocean, and (b) sent via satellite 36,000 km above the ocean. Would this cause a noticeable delay in either case
Answer:
a) t = 2 10⁻² s , t = 2.4 10-1 s
Explanation:
In this exercise they indicate the delay in two signals
a) A signal travels on an electrical cable, between New York and London.
The wave formed in this wire for the signal. This wave travels at the speed of light, c = 3 108 m / s, so the delay is very small
t = d / c
t = 6000 10³/3 10⁸
t = 2 10⁻² s
b) The signal points to a satellite in geostationary orbit
distance traveled = √ (2 36000 10³)² + 6000²
distance = 72 10⁶ m
the waiting time is
t = d / c
t = 72 10⁶/3 10⁸
t = 2.4 10-1 s
we can see that the signal sent by the satellites has more delay because its distance is much greater
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.
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
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a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.
Explanation:
It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is
[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]
[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
A)The power output of the microwave oven is?
B) The intensity of the microwave beam is?
C) The electric field amplitude is?
D) The force on the base due to the radiation is?
Answer:
power = 600000 W
intensity = 6666666.66 W/m²
Em = 70880.18 N/m
F = 2 × [tex]10^{-3}[/tex] N
Explanation:
given data
frequency f = 2400 MHz
height oven cavity h = 25 cm = 0.25 m
base area measures A = 30 cm by 30 cm
total microwave energy content of cavity E = 0.50 mJ = 0.50 × [tex]10^{-3}[/tex]
solution
first, we get here total time taken from top to bottom that is express as
Δt = [tex]\frac{h}{c}[/tex] ...............1
Δt = [tex]\frac{0.25}{3\times 10^8}[/tex]
Δt = 8.33 × [tex]10^{-10}[/tex] s
and
power output will be
power = [tex]\frac{E}{\Delta t}[/tex] ..............2
power = [tex]\frac{0.50 \times 10^{-3}}{8.33 \times 10^{-10}}[/tex]
power = 600000 W
and
intensity of the microwave beam is
intensity = power output ÷ base area ..............2
intensity = [tex]\frac{600000}{30 \times 30 \times 10^{-4}}[/tex]
intensity = 6666666.66 W/m²
and
electric field amplitude is
as we know intensity I = [tex]\frac{E^2}{c \mu o}[/tex] ...............3
[tex]E(rms) = \sqrt{Ic\ \mu o} \\E(rms) = \sqrt{6666666.66 \times 3 \times 10^{8} \times 4 \pi \times 10^{-7} }[/tex]
E(rms) = 50119.87 N/m
and we know
[tex]E(rms) = \frac{Em}{\sqrt{2}}\\50119.87 = \frac{Em}{\sqrt{2}}[/tex]
Em = 70880.18 N/m
and
force on the base due to the radiation is by the radiation pressure
[tex]Pr = \frac{l}{c}[/tex] ..................4
[tex]\frac{F}{A} = \frac{l}{c}[/tex]
so
F = [tex]\frac{6666666.66 \times 900 \times 10^{-4}}{3\times 10^8}[/tex]
F = 2 × [tex]10^{-3}[/tex] N
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
Water has a specific heat capacity of 1.00 cal/g °C, and copper has a specific heat capacity of 0.092 cal/g °C. If both are heated to 100 °C, which takes longer to cool?
Answer:
The water takes longer, because it is the better insulator here.
Explanation:
Conductors and insulators work similarly in "reverse".
If something is a good heat conductor, then it's good at both absorbing heat energy and giving it away. Insulators are good at resisting temperature changes, but also take longer to cool down once they are heated up.
So because copper is the better conductor here, it will cool faster than the water at the same temperature.
Which of the following describes wavelength?
A.
the height of a wave
B.
the distance between crests of adjacent waves
C.
the distance a wave travels in a given amount of time
D.
the number of waves that pass a point in a given amount of time
The positron has the same mass as an electron, with an electric charge of +e. A positron follows a uniform circular motion of radius 5.03 mm due to the force of a uniform magnetic field of 0.85 T. How many complete revolutions does the positron perform If it spends 2.30 s inside the field? (electron mass = 9.11 x 10-31 kg, electron charge = -1.6 x 10-19 C)
Answer:
5.465 × 10^10 revolutions
Explanation:
Formula for Magnetic Field = m. v/ q . r
M = mass of electron = mass of positron = 9.11 x 10^-31 kg,
radius of the positron = 5.03 mm
We convert to meters.
1000mm = 1m
5.03mm = xm
Cross multiply
x = 5.03/1000mm
x = 0.00503m
q = Electric charge = -1.6 x 10^-19 C
Magnetic field (B) = 0.85 T
Speed of the positron is unknown
0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503
0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v
v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg
v = 6.8408 ×10-22/ 9.11 x 10^-31 kg
v = 750911086.72m/s
Formula for complete revolutions =
Speed × time / Circumference
Time = 2.30s
Circumference of the circular path = 2πr
r =0.00503
Circumference = 2 × π × 0.00503
= 0.0316044221
Revolution = 750911086.72 × 2.30/0.0316044221
= 1727095499.5/0.0316044221
= 546541562294 revolutions
Approximately = 5.465 × 10^10 revolutions
The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V.
A) What is the frequency of the output voltage?
B) What is the strength of the magnetic field in which the coil is turning?
Answer:
A) 60 Hz
B) 0.04 T
Explanation:
Given that.
Number of turns, N = 200
Length of the side, l = 20 cm = 0.2 m
Speed if rotation, w = 3600 rpm
Voltage, V = 120 V
First, we try to convert the speed from rpm to rad/s
3600 * (2π/60)
3600 * 0.10473
3600 rpm = 377 rad/s
Now, we use that as our w, speed of rotation
Frequency of output, f =
w/2π
f = 377 / 6.284
f = 59.99 Hz or approximately, 60 Hz.
B
Strength of the magnetic field in which the coil is turning
E• = NABw
Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have
120 = 200 * 0.04 * 377 * B
120
Making B subject of formula,
B = 120/ 3016
B = 0.04 T..
The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above water when his lungs are full.
Required:
a. Calculate the volume of air he inhales - called his lung capacity - in liters.
b. Does this lung volume seem reasonable?
Answer:
A) V_air = 1.295 L
B) Volume is not reasonable
Explanation:
A) Let;
m be total mass of the man
m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung
m_3 be the mass above the water with the empty lung
m_5 be the mass above the water with full lung
F_b be the buoyant force due to the air in the lung
V_a be the volume of air inside man's lungs
w_p be the weight that the buoyant force opposes as a result of the air.
Now, we are given;
m = 78.5 kg
m_3 = 3.2% × 78.5 = 2.512 kg
m_5 = 4.85% × 78.5 = 3.80725 kg
Now, m_p = m_5 - m_3
m_p = 3.80725 - 2.512
m_p = 1.29525 kg
From archimedes principle, we have the formula for buoyant force as;
F_b = (m_displaced water)g = (ρ_water × V_air × g)
Where ρ_water is density of water = 1000 kg/m³
Thus;
F_b = w_p = 1.29525 × 9.81
F_b = 12.7064 N
As earlier said,
F_b = (ρ_water × V_air × g)
Thus;
V_air = F_b/(ρ_water × × g)
V_air = 12.7064/(1000 × 9.81)
V_air = 1.295 × 10^(-3) m³
We want to convert to litres;
1 m³ = 1000 L
Thus;
V_air = 1.295 × 10^(-3) × 1000
V_air = 1.295 L
B) From research, the average lung capacity of an adult human being is 6 litres of air.
Thus, the calculated lung volume is not reasonable
What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still water level?
Answer:
physical feature of a wave is related to the depth of the wave base is The circular orbital motion
B. The wave base is the depth, and the still water level is the horizontal level
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 trombone can be modeled like an open closed air tube. the trombone plays a fifth harmonic of 159 hz. the speed of sound is 343 m/s. what is the wavelength of this sound
Answer:
The wavelength is 2.16 m.
Explanation:
Given the speed of the sound = 343 m/s
Trombone generate the frequency = 159 Hz
Now we have to find the wavelength of the sound. Here, we can find the wavelength by dividing the speed of the sound with frequency.
The wavelength of the sound = Speed of sound/frequency
Wavelength of the sound = 343 / 159 = 2.16 m
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:
a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?
Answer:
8.78 AmpsExplanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
Hence the current when the unit/heater is operating is 8.78 AmpAn intergalactic rock star bangs his drum every 1.30 s. A person on earth measures that the time between beats is 2.50 s. How fast is the rock star moving relative to the earth
Answer:
v = 0.89 c = 2.67 x 10⁸ m/s
Explanation:
The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:
t = t₀/√[1 - v²/c²]
where,
t = time measured by the person on earth = 2.50 s
t₀ = rest time of the intergalactic rock star = 1.30 s
v = relative speed of the rock star = ?
Therefore,
2.5 s = (1.3 s)/√[1 - v²/c²]
√[1 - v²/c²] = 1.3/2.5
√[1 - v²/c²] = 0.52
[1 - v²/c²] = 0.52²
[1 - v²/c²] = 0.2074
v²/c² = 1 - 0.2074
v²/c² = 0.7926
v/c = √0.7926
v = 0.89 c
where,
c = speed of light = 3 x 10⁸ m/s
v = (0.89)(3 x 10⁸ m/s)
v = 0.89 c = 2.67 x 10⁸ m/s
The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement
Explanation:
1 inch = 25.4 mm
1 foot = 12 inches
1 mile = 5260 feet
1 cm = 0.01 m or 10 mm
Now Sammy's height is 5 feet and 5.3 inches.
(a) We need to find Sammy's height in inches.
Since, 1 foot = 12 inches
5 feet = 5 × 12 inches = 60 inches
Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches
Sammy's height is 65.3 inches.
(b) We need to find Sammy's height in feet.
Since, 1 foot = 12 inches
[tex]1\ \text{inch}=\dfrac{1}{12}\ \text{feet}[/tex]
So,
[tex]5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}[/tex]
5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet
Sammy's height is 5.44 feet.
Calculate the equivalent of 30 degrees Celsius and 50 degrees Celsius on a Kelvin
[tex]\boxed{\sf 1°C=273K}[/tex]
Sol:-1
[tex]\\ \sf\longmapsto 30°C[/tex]
[tex]\\ \sf\longmapsto 273+30[/tex]
[tex]\\ \sf\longmapsto 303K[/tex]
Sol:-2
[tex]\\ \sf\longmapsto 50°C[/tex]
[tex]\\ \sf\longmapsto 50+273[/tex]
[tex]\\ \sf\longmapsto 323K[/tex]
A metal blade of length L = 300 cm spins at a constant rate of 17 rad/s about an axis that is perpendicular to the blade and through its center. A uniform magnetic field B = 4.0 mT is perpendicular to the plane of rotation. What is the magnitude of the potential difference (in V) between the center of the blade and either of its ends?
We are being given that:
The length of a metal blade = 300 cmThe angular velocity at which the metal blade is rotating about its axis is ω = 17 rad/sThe magnetic field (B) = 4.0 mTA pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.
From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).
The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:
[tex]\mathsf{E = B\times l\times v}[/tex]
where;
B = magnetic fieldl = lengthv = relative speedFrom the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.
Then, the velocity (v) = ωx
The potential difference across it can now be expressed as:
[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]
For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.
∴
[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]
[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]
[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]
Recall that,
magnetic field (B) = 4 mT = 4 × 10⁻³ TLength L = 300 cm = 3mangular velocity (ω) = 17 rad/s[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]
[tex]\mathsf{E = 19.13 mV}[/tex]
Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.
Learn more about Faraday's law of induction here:
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A ceiling fan is spinning at 45 revolutions per minute when it is switched to a higher speed. It accelerates uniformly, and 2.0 seconds later it is spinning at 110 revolutions per minute. Through how many radians did it rotate during the transition of speeds
Answer:
θ = 16.2 rad
Explanation:
First we find the angular acceleration by using first equation of motion in angular form:
ωf = ωi + αt
where,
ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s
ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s
α = angular acceleration = ?
t = time = 2 s
Therefore,
11.5 rad/s = 4.7 rad/s + α(2 s)
α = (6.8 rad/s)/(2 s)
α = 3.4 rad/s²
Now, we use 2nd equation of motion:
θ = ωi t + (1/2)αt²
where,
θ = rotation = ?
Therefore,
θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²
θ = 9.4 rad + 6.8 rad
θ = 16.2 rad
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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How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F
Answer:
Explanation:
To convert from °C to °F , the formula is
( F-32 ) / 9 = C / 5
F is reading fahrenheit scale and C is in centigrade scale .
F = 205 , C = ?
(205 - 32) / 9 = C / 5
C = 96°C approx .
Let us calculate the heat required .
Total heat required = heat required to heat up the ice at - 20 °C to 0°C + heat required to melt the ice + heat required to heat up the water at 0°C to
96°C.
= 5 x 2.04 x (20-0) + 5 x 336 + 5 x ( 96-0 ) x 4.2 kJ .
= 204 + 1680 + 2016
= 3900 kJ .
Does the moon light originate from the moon only
Answer:
No
Explanation:
Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.
:-)