Answer:
The depth will be equal to 6141.96 m
Explanation:
pressure on the submarine [tex]P_{sea}[/tex] = 62 MPa = 62 x 10^6 Pa
we also know that [tex]P_{sea}[/tex] = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
[tex]P_{sea}[/tex] = 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine [tex]P_{g}[/tex] = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure [tex]P_{sea}[/tex], we have
62 x 10^6 = 10094.49h
depth h = 6141.96 m
Write the name of the law for each of the equations and explain in one or two lines how the two sides are related maxwells equation.
Answer:
Lorenz force law
Explanation:
The equation given name after the physicist and the mathematician James Clerk Maxwell. He published his first equation in 1861 and 1862 that included the Lorenz force law. These equation were used by Maxwell first of all for the light in electromagnetic phenomena.
In his equations Maxwell proposed that how the electric and magnetic field is fluctuate and constant speed in the vacuum which is called a electromagnetic radiation.
These are waves that occurs at different level of wavelength spectrum. These equation have two major variant such as universal applicability. The Maxwell equations are used for the Equivalent alternative formulation.
When asked how to create an electromagnet, the best answer would be:
"You can create an electromagnet by applying a direct current to a non–magnetic material in order to create an electromagnetic field."
"You can create an electromagnet by covering a wire with insulation and applying a direct electrical current to the magnet."
"You can create an electromagnet by coiling a ferromagnetic material, wrapping in an insulator, and applying a current to the coil."
"You can create an electromagnet by wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current."
Answer:
You can create an electromagnet by wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current."
Explanation:
Electromagnets are made by wrapping an insulated wire around a metal with ferromagnetic properties (example is iron), to form a loop, and then applying a current through the wire. Electromagnets can generate magnetism with a strong force field, and unlike normal magnets, their strength can be varied by varying the amount of current flowing through the coil. Their main disadvantage, which is also their most utilized property is that their magnetism is lost once the current flowing through the wire is cut-off.
Answer:
You can create an electromagnet by wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current."
Explanation:
A square coil with a side length of 16.0 cm and 29 turns is positioned in a region with a horizontally directed, spatially uniform magnetic field of 83.0 mT and set to rotate about a vertical axis with an angular speed of 1.20 ✕ 102 rev/min.
(a) What is the maximum emf induced in the spinning coil by this field?
___V
(b) What is the angle between the plane of the coil and the direction of the field when the maximum induced emf occurs? (Enter the angle with the smallest possible magnitude.)
___°
Answer:
A
[tex]\epsilon_{max} = 0.774 \ V[/tex]
B
[tex]wt = 0^o[/tex]
Explanation:
From the question we are told that
The length of the side is [tex]l = 16.0 \ cm = 0.16 \ m[/tex]
The number of turns is [tex]N = 29 \ turns[/tex]
The magnetic field is [tex]B = 83.0 mT = 83 *10^{-3} \ T[/tex]
The angular speed is [tex]w = 1.20 * 10^2 rev/min = \frac{1.20 *10^2 * 2\pi}{60 } = 12.6 \ rad/s[/tex]
Generally the area is [tex]A = l^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = N * w * B * A * cos(wt)[/tex]
At maximum [tex]cos(wt) = 1[/tex]
So
[tex]\epsilon_{max} = N * w * B * A[/tex]
[tex]\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* (l^2)[/tex]
=> [tex]\epsilon_{max} = 29 * 12.6 * 83*10^{-3}* ((0.16)^2)[/tex]
=> [tex]\epsilon_{max} = 0.774 \ V[/tex]
At maximum emf
[tex]cos (wt) = 1[/tex]
=> [tex](wt) = cos^{-1} (1)[/tex]
=> [tex]wt = 0^o[/tex]
A 76-W incandescent light bulb operates at 120 V. How many electrons and coulombs flow through the bulb in one day?
Answer:
43200c
Explanation:
You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of polished metal with a 44 cm radius of curvature.
(a) Where is the image of your 5.0-cm-tall nose located?
(b) What are the image’s size, orientation, and nature (real or virtual)?
Answer:
a) q = 39.29 cm , b) h ’= - 3.929 cm the image is inverted and REAL
Explanation:
For this exercise we will use the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length of the salad bowl, p and q are the distance to the object and the image
The metal salad bowl behaves like a mirror, so its focal length is
f = R / 2
f = 44/2
f = 22 cm
a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image
1 / q = 1 / f - 1 / p
1 / q = 1/22 - 1/50
1 / q = 0.0254
q = 39.29 cm
b) to calculate the size of the image we use the equation of magnification
m = h’/ h = - q / p
h ’= - q / p h
h ’= - 39.29 / 50 5
h ’= - 3.929 cm
the negative sign means that the image is inverted
as the rays of light pass through the image this is REAL
PLEASEE I NEED HELP FAST!!! .Study the scenario.A small container of water with a low temperature is poured into a large container of water with a higher temperature. Which choice correctly explains what happens to the thermal energy of these systems? A)The smaller container of water has more thermal energy than the larger container, and some of that energy is transferred to the warmer water in a process known as heating. B)The larger container of water has more thermal energy and some of that energy is transferred to the colder water in a process known as heating. C)The larger container of water has more heat and thermal energy than the smaller container. Some heat and thermal energy is transferred to the smaller container of water. D)The smaller container of water has more heat and thermal energy than the larger container. Some heat and thermal energy is transferred to the larger container of water.
Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce ([tex]T_{f}[/tex]-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
The elements within a group tend to share
A. similar chemical properties and characteristics
B. similar atomic weights
c. similar atomic numbers
D. similar atomic symbols
Answer: A. similar chemical properties and characteristics
Let k be the Boltzmann constant. If the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value, the entropy of the gas changes by:__________. A) S = 3kln2 B) S =
Answer:
ΔS = - k ln (3)
Explanation:
Using the Boltzmann's expression of entropy, we have;
S = k ln Ω
Where;
S = Entropy
Ω = Multiplicity
From the question, the configuration of the molecules in a gas changes so that the multiplicity is reduced to one-third its previous value. This also causes a change in the entropy of the gas as follows;
ΔS = k ln (ΔΩ)
ΔS = kln(Ω₂) - kln(Ω₁)
ΔS = kln(Ω₂ / Ω₁) -------------(i)
Where;
Ω₂ = Final/Current value of the multiplicity
Ω₁ = Initial/Previous value of the multiplicity
Ω₂ = [tex]\frac{1}{3}[/tex] Ω₁ [since the multiplicity is reduced to one-third of the previous value]
Substitute these values into equation (i) as follows;
ΔS = k ln ([tex]\frac{1}{3}[/tex] Ω₁ / Ω₁)
ΔS = k ln ([tex]\frac{1}{3}[/tex])
ΔS = k ln (3⁻¹)
ΔS = - k ln (3)
Therefore, the entropy changes by - k ln (3)
A beam of light in air strikes a slab of glass (n = 1.52) and is partially reflected and partially refracted and partially refracted, find the angle of incidence if the angle of reflection is twice the angle of refraction.
θ1= (answer in degress)
Answer:
Explanation:
angle of reflection = angle of incidence i = 2θ
angle of refraction r = θ
sin i / sin r = 1.52
sin2θ / sin θ = 1.52
2 sinθ . cosθ / sin θ = 1.52
2 cosθ = 1.52
cosθ = .76
θ= 41°
Hence angle of incidence = 41°
.
An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 43.0 km . To the dismay of scientists on earth, an electrical fault causes an on-board thruster to fire, decreasing the speed of the spacecraft by 23.0 m/s .If nothing is done to correct its orbit, with what speed (in km/h) will the spacecraft crash into the lunar surface?
Answer: v₂ = 5962 km
the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
Explanation:
Given that;
Lunar surface is in an altitude h = 43.0 km = 43 × 10³ m
we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²
speed of the space craft when it crashes into the lunar surface , v
decreasing speed of the space craft = 23 m/s
Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r
but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²
HERE r = Rm + h
we substitute
r = 1.74 × 10⁶ m + 43 × 10³ m
= 1.783 × 10⁶ m
On equating these, we have
G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)
v²/r = GM/r²
v = √ ( GM/r)
v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )
v = √ (2750787.9978)
v = 1658.55 m/s
Now since speed is decreasing by 23 m/s
the speed of the space craft into the lunar face is,
v₁ = 1658.55 m/s - 23 m/s
v₁ = 1635.55 m/s
Now applying conversation of energy, we say
1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)
v₂ = √ [ v₁² + GMe (1/Rm - 1/r)]
v₂ = √ [ 1635.55² + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]
v₂ = √ (2675023.8025 + 67979.24)
v₂ = √(2743003.046)
v₂ = 1656.2 m/s
now convert
v₂ = 1656.2 × 1km/1000m × 3600s/1hrs
v₂ = 5962 km
Therefore the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
A coiled telephone cord forms a spiral with 52.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.5 cm. Determine the inductance of one conductor in the unstretched cord.
Answer:
The inductance of one conductor in the unstretched cord is 0.7849 μH
Explanation:
Given;
number of turns of the coil, N = 52 turns
diameter of the coil, d = 1.30 cm
radius of the coil, r = d/2 = 1.30 cm / 2 = 0.65 cm = 0.0065 m
length of the unstretched cord, l = 57.5 cm = 0.575 m
The inductance of one conductor in the unstretched cord is given by;
[tex]L = \frac{N^2 \mu_o A}{l}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
A is area of coil, = πr² = π x (0.0065)² = 1.328 x 10⁻⁴ m²
[tex]L = \frac{(52)^2(4\pi *10^{-7})(1.328*10^{-4})}{0.575} \\\\L = 7.849*10^{-7} \ H[/tex]
L = 0.7849 μH
Therefore, the inductance of one conductor in the unstretched cord is 0.7849 μH
A resistor has four colored stripes in the following order: orange, orange, brown and silver. What is the resistance of the resistor and its tolerance
Answer:
Resistance =330 Ω
Tolerance = 33 Ω
Explanation:
see attached resistor color code table
The first stripe is orange, which means the leftmost digit is a 3.
The second stripe is orange , which means the next digit is a 3.
The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.
The resistance is:
orange-orange-brown= 330 Ω
The tolerance is:
The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.
silver is 10%
A 330 Ω resistor has a silver tolerance band.
Tolerance = value of resistor x value of tolerance band
= 330 Ω x 10% = 33 Ω
330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.
The resistance of the resistor is 330 Ω and the tolerance is within 363 Ω and 297 Ω
In physics, resistor's resistance is coded using colors.
Orange colors are coded as 3
The brown color is coded as 0
The silver color determines the tolerance and silver means 10%
The resistor with four colored stripes in the following order: orange, orange, brown has a resistance value of 330 Ω
Tolerance = 330 × 10%
Tolerance = 33Ω
Resistor value = 330±33
Resistor value = (330+33) and (330-33)
Resistor value = 363 Ω and 297 Ω
Hence the resistance of the resistor is 330 Ω and the tolerance is within 363 Ω and 297 Ω
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Which event would most likely occur if Earth did not retain the heat from its formation? The inner core would liquefy. Seismic waves would move the crust. The Earth’s magnetic field would disappear. The asthenosphere and outer core would solidify.
Answer:
D-The asthenosphere and outer core would solidify.
Explanation:
The athenosphere is a region of the earth mantle below the lithosphere.
If the Sun didn’t retain the heat formed during its formation then it will have an adverse effect on other parts such as the core and mantle.The inner core wouldn’t liquefy but solidify. The asthenosphere and outer core would also solidify.
The Seismic waves wont be able to move the crust and the Earth’s magnetic field wouldn’t disappear but would be more pronounced.
The Earth’s magnetic field would disappear as well as solidification of asthenosphere and outer core will occur.
The asthenosphere and outer core would solidify and the Earth’s magnetic field would disappear if the Earth did not retain the heat from its formation because this heat make all the materials present in the outer core and asthenosphere in liquid form. This liquid form in the outer core is responsible for the generation of magnetic field around the earth.
If this heat will not retain in asthenosphere and outer core, the heat will escape and cooldown the asthenosphere and outer core which change it into solid form so the Earth’s magnetic field will be disappear so we can conclude that the Earth’s magnetic field would disappear as well as solidification of asthenosphere and outer core will occur if the Earth did not retain the heat from its formation.
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The longest pipe found in most medium-size pipe organs is 2.40 m (7.87 ft) long. What is the frequency of the note corresponding to the fundamental mode if the pipe is open at one end and closed at the other
Answer:
34.3Hz
Explanation:
We know that
For a pipe open at one end, fundamental frequency (Fo)
is
Fo = v/4L
Where v= 340m/s
Fo = 340/4 x 2.48 = 34.3Hz
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Answer:
[tex]f=1.98\times 10^5\ Hz[/tex]
Explanation:
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by :
[tex]B=1.2\times 10^{-6}\sin [2\pi (\dfrac{z}{240}-\dfrac{10^7t}{8})][/tex] .....(1)
The general equation of the magnetic field wave is given by :
[tex]B=B_o\sin (kz-\omega t)[/tex] ....(2)
Equation (1) is in form of equation (2), if we compare equation (1) and (2) we find that,
[tex]\omega=\dfrac{10^7}{8}[/tex]
We need to find the frequency of the wave. It is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{10^7}{8\times 2\pi}\\\\f=1.98\times 10^5\ Hz[/tex]
So, the frequency of the wave is [tex]1.98\times 10^5\ Hz[/tex]
A man in a boat is lookinh dtraight down at a fish in the water (n = 1.333) directly beneath him. The fish is looking straight up at the man. They are equidistant from the air/water interface. To the man, the fish appears to be 2.3 m beneath his eyes. To the fish, how far above its eyes does the man appear to be?
Answer:
To the fish the man appears to be 3.06m above its eyes
Explanation:
We know that refractive index n
n = real dept/ apparent depth
While apparent depth is the distance the fish appears to the man which is 2.3m so using the equation
Real dept= n x apparent depth
= 1.333* 2.3= 3.06m
Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is exactly replaced by the second-order maximum of B. (a) Determine the ratio dB/dA (No units) of the spacings between the slits of the gratings. Then find the next two principla maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched [(b) and (c) for the first pair, (d) and (e) for the second pair]. Identify these maxima by their order numbers (No units), listing the maxima for grating B first.
Answer:
a) dB / dA = 2 ,
b) Network B Network A
2 1
4 2
6 3
Explanation:
a) The expression for grating diffraction is
d sin θ = m λ
where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range
In this exercise we are told that the two spectra are in the same position, let's write the expression for each network
Network A
m = 1
sin θ = 1 λ / dA
Network B
m = 2
sin θ = 2 λ / dB
they ask us for the relationship between the distances, we match the equations
λ / dA) = 2 λ / dB
dB / dA = 2
b) let's write the equation of the networks
sin θ = m_A λ / dA
sin θ = m_B λ / dB
we equalize
m_A λ/ dA = m_B λ / dB
we use that
dB / dA = 2
m_A 2 = m_B
therefore the overlapping orders are
Network B Network A
2 1
4 2
6 3
The uncertainty in position of a proton confined to the nucleus of an atom is roughly the diameter of the nucleus. If this diameter is d = 7.8×10−15 m, what is the uncertainty in the proton's momentum?
Answer:
0.135E-19kgm/s
Explanation:
Using the uncertainty principle, we find
Dp = h / (4π Dx)
= (6.63×10-34Js)/4π(3.90×10-15m) = 0.135×10-19kg m/s
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
The relationship is [tex]E=\frac{ kQ}{d^2}[/tex]
Explanation:
The electric field strength is denoted by the symbol E,
the test charge is denoted be q and the source charge be Q
distance is denoted by d
Then the equation can be rewritten in symbolic form as
Electric field strength is = Force/charge
[tex]E=\frac{F}{q}------1[/tex]
we know that the formula for force is given as
[tex]F= \frac{kQq}{d^2} ------2[/tex]
where [tex]k= 9*10^9 N.m^2/C^2[/tex]
and d is the separation distance between charges
We can insert the expression for Force in equation one
we have
[tex]E= \frac{kQ\sout{q}/d^2}{\xout{q}}--------3[/tex]
We can strike out both qs in the numerator and denominator we have
[tex]E=\frac{ kQ}{d^2}[/tex]
The electric field in a traveling em wave in the vacuum has an rms intensity of 6.75 v/m.
Required:
Calculate the amount of energy delivered in average during 3.02x10^7s to 2.00 cm^2 of a wall that it hits perpendicularly?
Answer:
The amount of energy delivered is 730.84 J
Explanation:
Given;
rms intensity of the electric field, [tex]E_{rms}[/tex] = 6.75 v/m
area of the wall, A = 2.0 cm² = 2.0 x 10⁻⁴ m²
The average intensity of the wave is given by;
[tex]I_{avg} = c \epsilon_o E_{rms}^2[/tex]
where;
c is the speed of light
ε₀ is permittivity of free space
I = (3 x 10⁸)(8.85 x 10⁻¹²)(6.75)²
I = 0.121 W/m²
Average power delivered, P = I x A
P = 0.121 x 2 x 10⁻⁴
P = 2.42 x 10⁻⁵ W
The amount of energy delivered is calculated as;
E = P x t
E = 2.42 x 10⁻⁵ x 3.02 x 10⁷
E = 730.84 J
Therefore, the amount of energy delivered is 730.84 J
A truck covered 2/7 of a journey at an average speed of 40
mph. Then, it covered the remaining 200 miles at another
average speed. If the average for the whole journey was 35
mph, what was the amount of time for the whole journey?
h
Answer:
The amount of time for the whole journey is 8 hours.
Explanation:
A truck covered 2/7 of a journey at an average speed of 40 mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: [tex]1-\frac{2}{7} =\frac{5}{7}[/tex]
So if the truck covered the remaining 200 miles at [tex]\frac{2}{7}[/tex], this means that [tex]\frac{5}{7}[/tex] of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if [tex]\frac{5}{7}[/tex] of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?
[tex]miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles[/tex]
miles= 280
So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ([tex]speed=\frac{distance}{time}[/tex]), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:
[tex]time=\frac{distance}{speed}=\frac{280 miles}{35 mph}[/tex]
time= 8 h
The amount of time for the whole journey is 8 hours.
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of the wire? (The resistivity of gold is 2.44 × 10-8 Ω·m.) A. 9.0 μm B. 8.5 μm C. 17 μm D. 48 μm
Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, [tex]\rho=2.44\times 10^{-8}\ \Omega-m[/tex]
Resistance in terms of resistivity is given by :
[tex]R=\dfrac{\rho l}{A}[/tex]
Also, V = IR
So,
[tex]\dfrac{V}{I}=\dfrac{\rho l}{A}[/tex]
A is area of wire,
[tex]\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}[/tex], r is radius, r = d/2 (diameter=d)
[tex]\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m[/tex]
Out of four option, near option is (C) 17 μm.
The diameter of the wire is 18μm and the closest answer is option C.
The resistance of a wire is proportional to its length and inversely proportional to its area.
Given that a 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. Then,
R = (ρL)/A
where
ρ = resistivity = 2.44 × 10-8 Ω·m
L = length
A = Area = [tex]\pi ^{2}[/tex]D/2
D = diameter of the wire
From Ohm's law, V = IR
make resistance R the subject of the formula
R = V/I
R = 1.5/4 x [tex]10^{-3}[/tex]
R = 375 Ω
Substitute all the parameters into the formula
R = (ρL)/A
375 = (2.44 × [tex]10^{-8}[/tex] x 4)/A
A = (9.76 × [tex]10^{-8}[/tex])/ 375
A = 2.603 x [tex]10^{-10}[/tex] [tex]m^{2}[/tex]
but
A = [tex]\pi (D/2)^{2}[/tex]
2.603 x [tex]10^{-10}[/tex] = [tex]\pi (D/2)^{2}[/tex]
[tex]\pi[/tex][tex]D^{2}[/tex] /4 = 2.603 x [tex]10^{-10}[/tex]
[tex]\pi[/tex][tex]D^{2}[/tex] = 1.04 x [tex]10^{-9}[/tex]
[tex]D^{2}[/tex] = (1.04 x [tex]10^{-9}[/tex])/ [tex]\pi[/tex]
D = [tex]\sqrt{3.3 * 10^{-10} }[/tex]
D = 1.8 x [tex]10^{-5}[/tex] m
D = 18 μm
Therefore, the diameter of the wire is 18μm and the closest answer is option C.
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A nearsighted woman has a far point of 180 cm . Part A. What kind of lens, converging or diverging, should be prescribed for her to see distant objects more clearly?Part B. What power should the lens have?
Explanation:
It is given that,
A nearsighted woman has a far point of 180 cm.
The object distance for nearsightedness is infinity, u = ∞
The image distance is, v = -180 cm
Using lens formula,
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{f}=\dfrac{1}{(-180)}-\dfrac{1}{\infty}\\\\f=-180\ cm\\\\f=-1.8\ m[/tex]
For nearsightedness, a diverging lens is used.
Power of the lens,
P = 1/f
[tex]P=\dfrac{1}{-1.8}\\\\P=-0.56\ D[/tex]
So, the power of the lens is -0.56 D.
The power of lens to be prescribed is 3.5 D.
A person that is far sighted can see far objects clearly but not nearby objects. The farpoint of a normal eye is infinity. Far sightedness is corrected by the use of a convex lens. The near point of the normal eye is 25 cm.
We have that;
u = 25 cm
v = -180 cm
1/f = 1/u - 1/v
1/f = 1/25 - 1/180
1/f = 0.04 - 0.0056
f = 29 cm
Power of lens = 100/f = 100/29 = 3.5 D
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You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 94 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 71 Hz. Take the speed of sound in air to be v = 340 m/s.
Part (a). Find an equation for the speed of the sound source. (In this case, it is the speed of the train.)
Part (b). Find the numeric value, in meters per second, for the speed of the train.
Part (c). Find an equation for the frequency of the train whistle that you would hear if the train were not moving.
Part (d). Find the numeric value, in hertz, for the frequency of the train whistle that you would hear if the train were not moving.
Answer:
Kindly check explanation
Explanation:
Given the following :
As train approaches ; frequency, f1 = 94Hz
As train recedes; frequency, f2 = 71Hz
Speed of sound in air ; v = 340m/s
A) speed of sound source (speed of train) = vs
From doppler effect :
As the train recedes ;
f2 = fs [v / (v + vs)] - - - - (1)
As train approaches :
f1 = fs [v / (v - vs)] ----- (2)
To find vs equate (1) and (2)
fs [v / (v - vs)] = fs [v / (v + vs)]
f1/f2 = v / (v - vs) ÷ v / (v + vs)
f1 / f2 = v / (v - vs) × (v + vs) / v
f1 / f2 = (v + vs) / (v - vs)
Let f1 / f2 = f
f = (v + vs) / (v - vs)
f (v - vs) = v + vs
fv - fvs = v + vs
fv - v = vs + fvs
v(f - 1) = vs(1 + f)
v(f - 1) / (1 + f) = vs
B)
v(f - 1) / (1 + f) = vs
f = f1 / f2 = 94/71 = 1.32 Hz
340(1.324 - 1) / (1 + 1.324) = vs
vs = 340(0.324) / 2.324
vs = 110.16 / 2.324
vs = 47.40 m/s
C.) To calculate fs, frequency of train, substitute vs into our equation.
f2 = fs [v / (v + vs)]
Following our substitikn we obtain:
fs = (2f / (f + 1))f2
D)
fs = (2f / (f + 1))f2
fs = 2(1.324) / (1.324 +1)) × 71
fs = (2.648 / 2.324) × 71
fs = 1.1394148 × 71
fs = 80.898450
fs = 80.90 Hz
If you shine an ultraviolet light on the metal ball of a negatively charged electroscope, what will happen
Answer:
The electroscope will become discharged by loosing electrons in a phenomenon called photoelectric effect
Explanation:
The UV light knocks off loosely bounded electrons in the outer shell causing the metal electroscope to discharge
Which statement describes a force acting on an object?
A.a person sees a rolling ball toward him
B. A magnet pulls a nail toward it
C. A piece of charcoal burns in a grill.
D.A car is parked on the street
Answer: B. A magnet pulls a nail towards it
Explanation: Out of the list of answers, this is the only answer where a force is acting on an object. the act of the magnet pulling on the nail is the force, and the nail is the object.
What is the radius of a tightly wound solenoid of circular cross-section that has 180 turns if a change in its internal magnetic field of 3.0 T/s causes a 6.0 A current to flow? The resistance of the circuit that contains the solenoid is 17 Ω. The only emf source for the circuit is the induced emf. A. 0.54 m B. 0.043 m C. 0.25 m D. 0.014 m
Answer:
C. 0.25 m
Explanation:
Given;
current flow in the solenoid, I = 6.0 A
number of turns of the solenoid, N = 180 turns
the resistance of the circuit, R = 17 Ω
change in the magnetic field, dB/dt = 3.0 T/s
The emf of the circuit is given by;
V = IR
V = 6 x 17
V = 102 Volts
Magnitude of induced emf is given by;
[tex]E = N(\frac{dB}{dt} )A[/tex]
Where;
A is area of the solenoid
[tex]E = N(\frac{dB}{dt} )A\\\\A = \frac{E}{N(\frac{dB}{dt} )} \\\\A = \frac{102}{180 *3} \\\\A = 0.18889 \ m^2[/tex]
Area of the circular solenoid is given by;
A = πr²
where;
r is radius of the solenoid
[tex]r = \sqrt{\frac{A}{\pi} }\\\\r = \sqrt{\frac{0.18889}{\pi} }\\\\r = 0.25 \ m[/tex]
Therefore, the correct option is C. 0.25 m
A 12.0-g sample of carbon from living matter decays at the rate of 162.5 decays/minute due to the radioactive 14C in it. What will be the decay rate of this sample in 1000 years? What will be the decay rate of this sample in 50000 years?
Answer:
a)143.8 decays/minute
b)0.41 decays/minute
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of C-14= 5670 years
t= time taken to decay
Ao= activity of a living sample
A= activity of the sample under study
a)
0.693/5670 = 2.303/1000 log(162.5/A)
1.22×10^-4 = 2.303×10^-3 log(162.5/A)
1.22×10^-4/2.303×10^-3 = log(162.5/A)
0.53 × 10^-1 = log(162.5/A)
5.3 × 10^-2 = log(162.5/A)
162.5/A = Antilog (5.3 × 10^-2 )
A= 162.5/1.13
A= 143.8 decays/minute
b)
0.693/5670 = 2.303/50000 log(162.5/A)
1.22×10^-4 = 4.61×10^-5 log(162.5/A)
1.22×10^-4/4.61×10^-5 = log(162.5/A)
0.26 × 10^1 = log(162.5/A)
2.6= log(162.5/A)
162.5/A = Antilog (2.6 )
A= 162.5/398.1
A= 0.41 decays/minute
Convert 50km/hr into m/s using dimensional analysis?
Answer:
Explanation:
1 km=1000 m
1 hour =60 min =60*60 sec=3600 sec
Now put 1000 m instead of km and 3600 sec instead of hour in the given expression.
=50 km/hour
=50*1000 m/3600 sec
=500 m/36 sec
=13.89 m/s
Answer:
= 13.89 m/s
Explanation:
km 1000 m 1 hr
50 ----- x ---------- x ------------
hr 1 km 3600 secs
= 13.89 m/s
Three identical point charges, Q = 2 PC, are placed at the vertices of an equilateral triangle as shown in the figure. The length of each side of the triangle is d
= 2 m. Determine the magnitude and direction of the total electrostatic force on the charge at the top of the triangle.
Answer:
The total force on the charge at the top of the triangle is approximately
[tex]1.56\,\,10^{-14}\,\,N[/tex]
Explanation:
Please look at the attached image to follow the explanation.
Since all charges are of the same positive value, the force exerted on the top charge by the other two are going to be on the line that joins the top charge with each of the other vertices of the triangle, pointing away from the top charge (illustrated by green vectors of the same length in the image).
We need to find the x and y components of these force vectors in order to find the resultant force by combining the x components among themselves, and the y components among themselves. Notice that the angle needed is in all cases [tex]60^o[/tex].
The x- components include the cosine of [tex]60^o[/tex], while the y components include the sine of [tex]60^o[/tex].
Notice as well that the x-components cancel each other (they have the same magnitude but point in opposite directions), while the y-components are both of the same value but pointing in the same direction (pointing up).
Then we just need to multiply by two the y component of one of the forces to find this total force.
Now, the magnitude of the forces in question are given by Coulomb's Law:
[tex]F_C=k\,\frac{q_1\,\,q_2}{d^2} =9\,\,10^9\,\frac{2\,\,10^{-12}\,\,2\,\,10^{-12}}{2^2} =9\,\,10^{-15}\,\,N[/tex]
therefore we can calculate what the y-component of this is using:
[tex]F_y=F\,sin(60^o)=9\,\,10^{-15} \,\frac{\sqrt{3} }{2} \\[/tex]
and twice this value becomes:
[tex]2\, \,F_y=(2)\,\,9\,\,10^{-15} \,\frac{\sqrt{3} }{2}=9\,\,\sqrt{3} \,\,10^{-15}\approx 1.56\,\,10^{-14}\,\,N[/tex]