Answer:
[tex]X=165 \textdegree[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda =360m[/tex]
Angle [tex]\theta=90 \textdegree[/tex]
Distance variation [tex]d=130m[/tex]
Generally the equation for phase difference X is mathematically given by
[tex]X=\theta+\frac{d}{\lambda}*100[/tex]
[tex]X=90+\frac{150}{360}*100[/tex]
[tex]X=165 \textdegree[/tex]
Therefore Phase difference X is given as
[tex]X=165 \textdegree[/tex]
8) Now drop the magnetic ball of the same size down the section of copper tube. Currents are induced in the tube both above and below the falling magnet (pictured here as a disc for clarity). On the drawing below, find the direction of the induced current, the direction of the induced magnetic field, and the N/S poles for both induced currents. Explain how these induced currents slow down the falling magnet.
Answer:
Explanation:
To have a deeper understanding of this. Let's begin with the acceleration due to gravity (g), with N pointing downward.
Since magnetic field lines originate in the north, field strength decreases for the upper disk and increases for the lower disk when the disk is in the center of two disks.
According to Lenz's theorem, current-induced would be so that the magnetic field due to this induced current opposes the magnetic field due to which there is an induced current, so induced current will be clockwise for the upper disk and anticlockwise for the lower disk.
If the south pole is pointing backward, the generated current will flow anticlockwise for the upper disk and clockwise for the lower disk.
Help me please please help
Answer:
The answer is C.
Explanation:
If you have ever tried doing so, hair stick to the ballon. Opposites attract as well, so the answer is C.
If the distance between the first order maximum and the tenth order maximum of a double-slit pattern is 18 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, what is the wavelength of the light used
Answer:
Wavelength of light is 600 nm
Explanation:
Given
Distance between the first order maximum and the tenth order maximum of a double-slit pattern = 18 mm
Separation between the slits = 0.15 mm
Distance of screen from the slits = 50 cm
Wavelength
[tex]= \frac{18*10^{-3} * 0.15 *10^{-3}}{0.50*9} \\= 6 *10^{-7}\\= 600[/tex]nm
How does the excretory system work with the circulatory system to help the body regulate itself?
Answer:
The excretory system is a close partner with both the circulatory and endocrine system. The circulatory system connection is obvious. Blood that circulates through the body passes through one of the two kidneys. Urea, uric acid, and water are removed from the blood and most of the water is put back into the system.
Explanation:
When electrons are accelerated by 2450 V in an electron microscope, they will have
wavelengths of:
a) 8.113 nm
b) 0.622 nm
c) 0.811 nm
The wavelength of the electron microscope is 2.472*10⁻²⁰ m.
Calculation:
Provided, accelerating voltage = 2450 V
According to de Broglie's equation:
λ = h/p
where,λ = the wavelength of the particle that moves
P= momentum of moving particle
h= plank's constant = 6.63 x 10⁻³ Js
We know that,
P = √(2mk)
P = √(2mqV)
where P= momentum of moving particle
M= mass of the moving object
K= kinetic energy of the object
V= Electric potential of the object(accelerating voltage)
q= charge of the object(electron) = e=1.6 x 10⁻¹⁹C
∴λ = h/√(2meV) = [tex]\frac{6,63 *10^{-34} }{2*9.1*10^{-31}*1.6*10^{-19}*2450 }[/tex]
=2.472*10⁻²⁰ m
S0, the wavelength of an electron microscope is 2.472*10⁻²⁰ m. Think it matches option C.
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HELP ME PLEASEEEEEEEEEEEEEE
Answer:
u r right
Explanation:
it is A. DNA
i took that class.
:)
Answer:
I think it may be a brain?
A ball is thrown downward from the top of a building with an initial speed of 25 m/s.
It strikes the ground after 2.0 s. How high is the building?
20 m
30 m
50 m
70 m
Answer:
h = 69.6 m
Explanation:
Data:
Vo = 25 m/st = 2.0 sg = 9.8 m/s²h = ?Formula:
[tex]\boxed{\bold{h=V_{0}*t+\frac{g*(t)^{2}}{2}}}[/tex]Replace and solve:
[tex]\boxed{\bold{h=25\frac{m}{s}*2.0\ s+\frac{9.8\frac{m}{s^{2}}*(2.0\ s)^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\frac{m}{s^{2}}+\frac{9.8\frac{m}{s^{2}}*4\ s^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+\frac{39.2\ m}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+19.6\ m}}[/tex][tex]\boxed{\boxed{\bold{h=69.6\ m}}}[/tex]The building has a height of 69.6 meters.
Greetings.
Does the mass change how a bouncy ball bounces
Answer:
hiii what's up. and good morning hyd
¿Cuál es la frecuencia de rotación en la tierra?
Answer:
frequency is approximately 11.5 µHz, or more exactly, 11.5740740e-6 Hz
Explanation:
if an object sinks in water its density is greater than that of water
The key insight that Bohr introduced to his model of the atom was that the angular momentum of the electron orbiting the nucleus was quantized. He introduced the postulate that the angular momentum could only come in quantities of nh/(2π), where h is Planck's constant and n is a nonnegative integer (0,1,2,3,…). Given this postulate, what are the allowable values for the velocity v of the electron in the Bohr atom? Recall that, in circular motion, angular momentum is given by the formula L= mvr.
Answer:
v = [tex]n \frac{\hbar }{m r}[/tex]
the sppedof the electron is also quantized
Explanation:
The angular momentum of a rotating body is
L = m v r
in Bohr's atomic model the quantization postulate is that the angular momentum is equal to
L = n [tex]\hbar[/tex]
we substitute
n [tex]\hbar[/tex] = m v r
v = [tex]n \frac{\hbar }{m r}[/tex]
where n is an integer.
Therefore, the sppedof the electron is also quantized, that is, sol has some discrete values.
A muon is a type of subatomic particle. If a muon is at rest in the laboratory, it will decay into an electron after about 2 microseconds. Suppose an observer watches a muon travel through the atmosphere at 90% of the speed of light. How does the lifetime of the moving muon compare to the laboratory muon for an observer at rest with respect to the lab
Answer:
[tex]\frac{t}{t_p}[/tex] = 2.29
Explanation:
For this exercise as the muon goes at speeds close to the speed of light we must use relativists
t =[tex]\frac{t_p}{\sqrt{1- (\frac{v}{c})^2 } }[/tex]
The proper time is the decay time in the reference frame where the muon is fixed ( laboratory), t_p = 2 10⁻⁶ s and the relation
v / c = 0.90
let's calculate
t = [tex]\frac{2 \ 10^{-6} }{\sqrt{1 \ - \ 0.9^2 } }[/tex]2 10-6 / Ra (1 - 0.9²)
t = 4.59 10⁻⁶ s
the ralation is
[tex]\frac{t}{t_p} = \frac{4.59 \ 10^{-6}}{ 2 \ 10^{-6}}\\[/tex]
[tex]\frac{t}{t_p}[/tex] = 2.29
The Sun is a star that Earth and the other planets revolve around. The Sun has a large gravitational pull. The gravitational attraction of the Sun contributes most to which of the following?
A) Motion of the planets
B) Magnetic field of the planets
C) Mass and density of the planets
D) Atmosphere of the planets
Answer:
A
Explanation:
The sun is an emits energy in the form of light and heat.
The planets absorb this energy which helps keep them in motion. The closer they are to the sun, the more energy, the more speed and less the time it takes a planet to orbit the sun.
The gravitational pull of the Sun contributes motion of the planets because here the gravitational pull of the sun is balanced by the centrifugal force produced by the revolution of planets. The correct option Is A.
What is centrifugal force?Centrifugal force is the pseudo-force that occurs only in the rotating frame of reference (non-inertial frame). it is directed away from the center. Its magnitude is,
F=[tex]\frac{mv^{2} }{r}[/tex]
where m= mass of the body in kg.
v= velocity f the body in m/s.
r = radius in m.
The star sun has a gravitational pull that is about 274 m/s² which is greater than the other planets. Due to this pull the planets tend to pull toward the sun but it does not happen because the pull force is balanced by the centrifugal force that is produced by the revolution of the planets around the sun.
Hence The Sun's gravitational pull contributes to the motion of the planets.
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You are working in the finance department of Innotech Ltd (INT). The Company has spent $5 million
in research and development over the past 12 months developing cutting-edge battery technology
which will be incorporated into the electric vehicle market. INT now need to choose between the
following three options for bringing the product to market. These options are:
Option 1: Manufacturing the product “in-house” and selling directly to the market
Option 2: Licensing another company to manufacture and sell the product in return for a royalty
Option 3: Sell the patent rights outright to the company mentioned in option 2
Your task
Your manager, INT’s CFO, Mr Barry Smith, has asked you to evaluate the three different options and
draft a memo to the Board of Directors providing recommendations on the alternatives, along with
supporting analyses.
Mr Smith has outlined the following three (3) areas you need to cover in your memo:
a) Analyse base case figures for the three options and using NPV as the investment decision
rule;
b) Provide recommendations based on the base-case analyses;
c) Provide recommendations on further analyses and discuss factors that should be considered
prior to making a final decision on the three options (Note. You do NOT have to undertake
any further financial analyses).
Further details for the various options are as follows:
Option 1: Manufacturing the product “in-house” and selling directly to the market
Three months ago, INT paid an external consultant $1.5 million for a production plan and demand
analysis. The consultant recommended producing and selling the product for five years only as
technological innovation will likely render the market too competitive to be profitable enough after
that time. Sales of the product are estimated as follows:
In the first year, it is estimated that the product will be sold for $45,000 per unit. However, the price
will drop in the following three years to $40,000 per unit and fall again to $36,000 per unit in the
final year of the project, reflecting the effects of anticipated competition and improving technology
Year Estimated sales volume
(units)
1 5,200
2 4,600
3 4,200
4 3,800
5 3,600
In the first year, it is estimated that the product will be sold for $45,000 per unit. However, the price
will drop in the following three years to $40,000 per unit and fall again to $36,000 per unit in the
final year of the project, reflecting the effects of anticipated competition and improving technology
in the market. Variable production costs are estimated to be $29,000 per unit for the entire life of the
project.
Fixed production costs (excluding depreciation) are predicted to be $3 million per year and marketing
costs will be $1.6 million per year.
Production will take place in factory space the company owns and currently rents to another business for $2.5 million per year. Equipment costing $87 million will have to be purchased. This
equipment will be depreciated for tax purposes using the prime cost method at a rate of 10% per
annum. At the end of the project, the company expects to be able to sell the equipment for $37
million.
Investment in net working capital will also be required. It is estimated that accounts receivable will
be 30% of sales, while inventory and accounts payable will each be 25% of variable and fixed
production costs (excluding depreciation). This investment is required from the beginning of the
project because credit sales, inventory stocks and purchases on trade credit will begin building up
immediately. All accounts receivable will be collected, suppliers paid and inventories sold by the end
of the project, thus the investment in net working capital will be returned at that point. (Refer to
spreadsheet example provided in Assessment Details).
Option 2: Licensing another company to manufacture and sell the product in return for a royalty
Lion Batteries Ltd (LIB), a multinational corporation, has expressed an interest in manufacturing and
marketing the pro
Answer: The Company has spent $5 million in research and development over the past 12 months developing cutting-edge battery technology which will be incorporated ...
Explanation: uhmmmmmm i dont know this one but it is pretty ez
A boy is pulling a sled with a net force of 10 N. If the mass of the sled is 20 kg, what is the acceleration of the sled?
Answer:
0.5 m/s
Explanation:
acceleration= force times mass
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
What is force?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
From Newton's Second Law, force is expressed as;
F = m × a
Where is mass of object and a is the acceleration
Given the data in the question;
Force applied F = 10N = 10kgm/s²Mass of the sled = 20kgAcceleration a = ?F = m × a
10kgm/s² = 20kg × a
a = 10kgm/s² ÷ 20kg
a = 0.5m/s²
Given the amount of force applied on the sled as well as its mass, the acceleration of the sled is 0.5m/s².
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Bobby Joe is having a fit and throws his backpack up in the air with an initial velocity of 26.2 m/s. Determine the height it will rise to before coming down and hitting Bobby Joe up the head.
A)2m
B)15m
C)25m
D)35m
Answer:
D) 35m"Explanation:
(and any subsequent words) was ignored because we limit queries to 32 words.In a stunt, three people jump off a platform and fall 8.5 m onto a large air bag. A fourth person at the other end of the air bag, the "flier," is launched 16 m vertically into the air. Assume all four people have a mass of 74 kg. What is the momentum of the flier just after launch? Let upward be the positive direction.
Answer:
They Died Period NO MORE
help Plz, these questions are super confusing .
Answer:
I don't know what the question is???????????????????????????????????????????????
While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 12 m, and you use your watch to find that each loop around takes 26 s.What is the magnitude of your acceleration
Answer:
a = 0.701 m / s²
Explanation:
As the game is spinning the acceleration is centripetal
a = v² /r
The park system after a small period of acceleration goes at a constant speed, for which we can use the relations of the uniform motion
v = d / t
the distance of a circle is
d = 2π r
we substitute
a = (2π r / t) ² / r
a = 4 pi² r / t²
let's calculate
a = 4 pi² 12 / 26²
a = 0.701 m / s²
A spring is hung from the ceiling. When a coffee mug is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation
Answer:
Explanation:
In equilibrium , weight of mug is equal to restoring force .
mg = kx where m is mass of mug , k is spring constant and x is extension .
k / m = g / x = 9.8 ms⁻² / .025 m
= 392
frequency of oscillation n = [tex]\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
[tex]n=\frac{1}{2\pi}\sqrt{392 }[/tex]
= 4.46 per second.
Pedro is planning to model how changes in weather affect evaporation from lakes for his first experiment he wants to test how humidity affects the evaporation rate. he places one beaker with 300 mL of water in a dry area. and places another beaker with 300 mL of water near a humidifier which of the following variables does Pedro need to control during his experiment
A. humidity only
B. humidity and evaporation rate
C. volume of water and tempature
D. volume of water only
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
What is meant by humidity ?The amount of water vapor in the air is known as humidity. The humidity will be high if there is a lot of water vapour in the atmosphere.
Water can evaporate even at very low temperatures, but as the temperature rises, the rate of evaporation increases.
More surface molecules per unit of volume may be able to escape from a substance with a larger surface area, so it will evaporate more quickly.
The control variables in an experiment are the variables that the experimenter intends to keep constant always so as to limit their effect on the measurements of the relationship between the dependent and the independent variable.
Therefore, in order to have a proper measurement of the effect of humidity on evaporation rate, other variables such as temperature, and the volume of the water in the experiment investigations which affect evaporation rate by the provision of heat, (temperature) and their heat capacity, the volume, etc. should be controlled.
Hence,
Pedro needs to control the variables such as volume of water and temperature during his experiment. So, option C.
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A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around the inner core. The spool is free to roll without slipping on a horizontal surface. If the string unwinds horizontally from the bottom of the core with a constant speed of 25.0 cm/s, what is the speed of the spool
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
[tex]\frac{v_o}{r_o} = \frac{v_1}{r_1}[/tex]
v1 = [tex]\frac{r_1}{r_o} \ \ v_o[/tex]
let's calculate
v₁ = [tex]\frac{1.50}{1.00} \ \ 25.0[/tex]
v₁ = 37.5 cm / s
94. The diagram shows the orbit of a satellite
around Earth. If Earth's mass is 5.97x10 kg,
what is the satellite's orbital speed?
(G=6.67x10-11 N·mº/kg?)
A 5.84x10' m/s
B 6.31x104 m/s
C 7.23x10 m/s
D 7.65x10 m/s
Answer:
7.65x10^3 m/s
Explanation:
The computation of the satellite's orbital speed is shown below:
Given that
Earth mass, M_e = 5.97 × 10^24 kg
Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg
Orbital radius, r = 6.80 × 10^6m
Based on the above information
the satellite's orbital speed is
V_o = √GM_e ÷ √r
= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6
= 7.65x10^3 m/s
⦁ Light coming from a fish makes an incidence angle of 30° to normal under the water. The index of refraction of water is 1.33. A fisherman is looking at the fish through air. At what angle with the normal will the fish appear to the fisherman? Use Snell’s law: .
Answer:
snell's law = sinI /sin r
refractive index = sinI /sin r
1.33= 30 /sinR
sinR= 1.33×30
sinR=39.9
The angle with the normal by which the fish appear to the fisherman is 41.68 degrees.
What is Snell's law?According to the Snell's law, the ratio of index of reflection of the different material is equal to the ratio of incident sine angle and reflective sine angle. It can be given as,
[tex]\dfrac{n_1}{n_2}=\dfrac{\sin\theta_2}{\sin\theta_1}[/tex]
Here (n₁ and n₂) are the index and reflective index and (θ₁ and θ₂) are the incident and reflected angle.
Light coming from a fish makes an incidence angle of 30° to normal under the water. The index of refraction of water is 1.33. A fisherman is looking at the fish through air.The index of refraction of air is 1. Thus, by the Snell's law,
[tex]\dfrac{1}{1.33}=\dfrac{\sin(30)}{\sin\theta_1}\\\theta_1=41.68^o[/tex]
Thus, the angle with the normal by which the fish appear to the fisherman is 41.68 degrees.
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A spring with a spring constant of 25.0N/m is stretched 4.50m. What is the force that the spring would apply?
Explanation:
[tex]from \: hookes \: law : \\ force = k \bold{.}e \\ = 25.0 \times 4.50 \\ = 112.5 \: N[/tex]
4. A 9000 kg train car travelling at 4.5 m/s inellastically collides with a 7500 kg train car at rest. If
all of the kinetic energy not conserved in this collision goes into increasing the internal energy of
the cars, how many Joules were absorbed?
Answer:
91125Joules
Explanation:
Given data
M1= 9000 kg
U1= 4.5m/s
M2= 7500 kg
U1= 0 m/s
KE1= 1/2M1U1^2
KE1= 1/2*9000*4.5^2
KE1= 1/2*9000*20.25
KE1= 1/2*182250
KE1=91125joules
KE2= 1/2M2U2^2
KE1= 1/2*7500*0^2
KE1=0 joules
Hence the energy absorbed is
E= KE1-KE2
E= 91125-0
E= 91125Joules
An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a resonance is heard when the water level is 180 cm below the top of the tube, and again after the water level is 220 cm below the top of the tube a resonance is heard. what is the frequency of the tuning fork? the speed of sound in air is 343 m/s. answer in units
Answer:
[tex]428.75\ \text{Hz}[/tex]
Explanation:
[tex]\Delta y[/tex] = Change in water level = [tex]220-180=40\ \text{cm}[/tex]
[tex]\lambda[/tex] = Wavelength
[tex]v[/tex] = Speed of sound = 343 m/s
Between the points of resonance there exists [tex]\dfrac{1}{2}\lambda[/tex]
[tex]\dfrac{1}{2}\lambda=\Delta y\\\Rightarrow \lambda=2\Delta y\\\Rightarrow \lambda=2\times 40\\\Rightarrow \lambda=80\ \text{cm}[/tex]
Wavelength is given by
[tex]f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{343}{0.8}\\\Rightarrow f=428.75\ \text{Hz}[/tex]
The frequency of the tuning fork is [tex]428.75\ \text{Hz}[/tex].
Light is traveling in glass, and hits a glass/unknown surface. In the glass the light beam is making an angle of 45.0 o with the normal to the surface. The glass has an index of refraction of 1.52. (A) If the refracted light ray leaves the glass at an 55o from the normal, what is the index of refraction for the unknown surface
Answer:
n = 1.31
Explanation:
When a ray of light crosses the separation surface between two transparent media, there exists a fixed relationship between the indexes of refraction of both media, related with the angles of incidence and refraction, which is known as Snell's Law.The Snell's Law can be written as follows:[tex]n_{i} * sin( \theta_{i}) = n_{r} * sin( \theta_{r}) (1)[/tex]
In our case the ray is incident from the glass, so ni = n glass = 1.52The angle of incidence is the angle that the ray makes with the normal to the separation surface, so θi=45º.The angle of refraction is the angle that the refracted ray makes with the normal, so θr= 55ºReplacing by the values in (1), and solving for nr, we have:[tex]n_{r} =\frac{n_{i} * sin \theta_{i} }{sin \theta_{r} } = \frac{sin (45)*1.52}{sin (55)} = 1.31 (2)[/tex]
Stars that are not very hot but give off a lot of light are
O nebula
O main sequence stars
giants
O giants
O dwarfs
[tex]\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}[/tex]
Giants.
thankshope it helps.Explanation:
giants are those stars that are not so hot but give a lot of light
Joshua's physics lab group has been assigned project of designing a model bobsled track Each group will roll a marble down the track with the objective of having the marbte finish with the fastest time. Assume that the track is frictionless, and the length of all tracks is the same. They can use a small marble with a mass of 15g or a large marble with a mass of 209. What should his team do to give their marble the most potential energy? A) They should use a small marble with the ramp at a 45° angle B) They should use a large marble with the ramp at a 60° angle, C) They should use a large marble with the ramp at a 75° angle. 2) They should use a small marble with the ramp at a 60° angle
They should use a large marble with the ramp at a 75 degree angle