Answer:
Yardangs are formed on vertical strata while zeugen on horizontal strata. ... Yardangs are formed on vertical hard/soft layers of rock, while zeugen (this is its plural form) are formed on horizontal bands of hard/soft rocks giving it a more mushroom-like shape. The Great Sphinx of Giza has been sculpted in a yardang
"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]
A car starts from rest and accelerates with a constant acceleration of 5 m/s2 for 4 s. The car continues for 18 s at constant velocity. How far has the car traveled from its starting point
A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 43 cmcm outside the aquarium.
Required:
What is the actual distance between the can and the aquarium?
Answer:
The actual distance is [tex]d_a = 0.3233\ m[/tex]
Explanation:
From the question we are told that
The distance of the can is d = 43 cm = 0.43 m
Generally the actual distance is mathematically represented as
[tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]
Where [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is
[tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]
So
[tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]
[tex]d_a = 0.3233\ m[/tex]
Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the speed of sound is 341 m/s, how fast is the object moving?
Answer:
The velocity is [tex]v = 8.743 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the signal sent out is [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]
The frequency of the signal received is [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]
The speed of sound is [tex]v_s = 341 \ m/s[/tex]
Generally the frequency of the sound received is mathematically represented as
[tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]
where v is the velocity of the object
=> [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]
=> [tex]1.05263 = \frac{341+v }{341-v}[/tex]
=> [tex]358.94 - 1.05263v = 341 + v[/tex]
=> [tex]17.947 = 2.05263 v[/tex]
=> [tex]v = 8.743 \ m/s[/tex]
An 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
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...
You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb?
a. at the point, because all rays bouncing off the mirror will be parallel.
b. at the focal point of the mirror
c. at the radius of curvature of the mirror
d. none of the above, you cant make parallel rays wilth a concave mirror
Answer:
Explanation:
Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.
For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.
Therefore, the light bulb should be placed at the focal point of the mirror.
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
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 .
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
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 flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
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]
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
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
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
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
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
A painter sets up a uniform plank so that he can paint a high wall. The plank is 2 m long and weighs 400 N. The two supports holding up the plank are placed 0.2 m from either end. Show that the upwards force on each of the planks is 200 N. Draw a sketch.
The upward force on each supporting plank is 200 N
The given parameters include;
weight of the plank, W₁ = 400 Nlength of the plank, l = 2 mupward force of each supporting plank, = W₂ and W₃To show that the upward force of each supporting plank is 200 N, make the following sketch.
W₂ W₃
↑ ↑
-----------------------------------------------------------------------
0.2m ↓ 0.2m
400 N
The two supporting planks keeps the 2m plank in equilibrium position. If the plank is in equilibrium position the sum of the upward forces equals sum of the downward force.W₂ + W₃ = 400 N
But the distance of each supporting plank from the end is equal, (0.2m).
Then, W₂ = W₃
2W₂ = 400 N
W₂ = 400N/2
W₂ = 200 N
W₃ = 200 N
Therefore, the upward force on each supporting plank that keeps the plank in equilibrium position is 200 N.
To learn more about equilibrium forces visit: https://brainly.com/question/12582625
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 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.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
A laser emits photons having an energy of 3.74 × 10–19 J. What color would be expected for the light emitted by this laser? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J ⋅ s)
Answer:
The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Explanation:
Given;
energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 x 10⁻³⁴ J.s
The wavelength of the emitted light will be calculated by applying energy of photons;
[tex]E = hf[/tex]
where;
E is the energy emitted light
h is Planck's constant
f is frequency of the emitted photon
But f = c / λ
where;
λ is the wavelength of the emitted photons
[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]
λ ≅ 532 nm
the wavelength of the emitted photons is 532 nm.
Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
"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.
You need to make a spring scale to measure the mass of objects hung from it. You want each 1.0 cm length along the scale to correspond to a mass difference of 0.10 kg. What should be the value of the spring constant?
Answer:
The spring constant should be:
[tex]k= 98\, \frac{N}{m}[/tex]
Explanation:
Use Hooke's law for this problem, knowing that the magnitude of the force (F) on the spring equals the stretching it experiences [tex]\Delta x[/tex] times the spring constant "k":
[tex]F=k\,\Delta x[/tex]
in our case, since the mass hanging is given in kg, we need to multiply it by "g" to get the force exerted:
Then if we add to the spring in its relaxed state, a mass of 0.10 kg, and we want for that a displacement of 1 cm (0.01 m), then the value of the spring constant should be:
[tex]k=\frac{F}{\Delta x} \\k=\frac{9.8\,(0.1)}{0.01} \, \frac{N}{m} \\k= 98\, \frac{N}{m}[/tex]
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
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .
Requried:
What is the magnitude of the charge (in nC) on each bead?
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
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.
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.
:-)
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
To get more information about Resistance :
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A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?
Answer:
6 rad/s²
Explanation:
Sum the torques about the hinge.
∑τ = Iα
mg(L/2) = mL²/3 α
g/2 = L/3 α
α = 3g/(2L)
α = 3 (10 m/s²) / (2 × 2.50 m)
α = 6 rad/s²