Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".
Kinetic energyAccording to the question,
Speed of light, c = 3 × 10⁸ m/s
Wavelength, λ = 120 nm or,
= 1.2 × 10⁻⁷ m
Plank's Constant, h = 6.626 × 10⁻³⁴ J.s
Now,
The energy of photon will be:
→ E = [tex]\frac{hc}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]
= [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]
= 10.35 eV
By using Einstein's Photoelectric equation,
Energy of Photon = Work function + K.E
10.35 = 4.82 + K.E
K.E = 10.35 - 4.82
= 5.53 eV or,
= 8.85 × 10⁻¹⁹ J
Thus the response above is correct.
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A car is going 8 meters per second on an access road into a highway
and then accelerates at 1.8 meters per second squared for 7.2
seconds. How fast is it then going?
Answer:
20.96 m/s^2 (or 21)
Explanation:
Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.
At first, we know a car is going 8 m/s, that is its initial velocity.
Then, we know the acceleration, which is 1.8 m/s/s
We also know the time, 7.2 second.
Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.
(final velocity - initial velocity) = time * acceleration
final velocity = time*acceleration + initial velocity
After plugging the found values in, we get 20.96 m/s/s, or 21 m/s
An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars
Answer:
Star A has a higher surface temperature than star B.
Explanation:
The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:
[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)
Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.
A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.
Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.
Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb
Answer:
300 W
Explanation:
power of each bulb P = 75 W
voltage in the circuit = 120 V
we know that electrical power P = IV ....1
and V = IR
we can also say that I = V/R
substituting for I in equation 1, we have
P = [tex]V^{2}/R[/tex] ....2
The total total power in the circuit = 75 x 2 = 150 W
from equation 2, we have
150 = [tex]120^{2} /R[/tex]
R = [tex]120^{2}/150[/tex] = 96 Ω this is the resistance of the whole circuit.
This resistance is due to the two light bulbs, for each light bulb since they are arranged in series
R = 96/2 = 48 Ω
From P = [tex]V^{2}/R[/tex]
for each light bulb, power is
P = [tex]120^{2} /48[/tex] = 300 W
PLEASE ANSWER ASAP
What happens to the ocean water before the precipitation part of the water cycle? ANSWERS; A.The ocean water condenses into the clouds. B.The ocean water collects back in the ocean. C.The ocean water falls back to Earth's surface. D. The ocean water runs off Earth's surface.
Answer:
B.
Explanation:
The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.
The ocean water prior to the part of the water cycle should be option B.
Ocean water:The ocean water should be collected back in the ocean prior to the part of the water cycle.
Because this should be done when it is evaporated by the sun. When the evaporation is done so the water should be transformed into water vapor.
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If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.
Required:
a. What is its frequency?
b. What type of electromagnetic radiation might this be?
Answer:
a) 5.5×10^17 Hz
b) visible light
Explanation:
Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;
λ= 5.5 × 10^-10 m
Since;
c= λ f and c= 3×10^8 ms-1
f= c/λ
f= 3×10^8/5.5 × 10^-10
f= 5.5×10^17 Hz
The electromagnetic wave is visible light
A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work. This heat engine violates the second law but not the first law of thermodynamics. Why is this true?
Answer:
It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J
But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law
A ball travels with velocity given by [21] [ 2 1 ], with wind blowing in the direction given by [3−4] [ 3 −4 ] with respect to some co-ordinate axes. What is the size of the velocity of the ball in the direction of the wind?
Answer:
2/5 m/s
Explanation:
There are two vectors v and w . Let θ be angle b/w the two vector.
[tex]cos\theta =\frac{\overleftarrow{v}\cdot \overleftarrow{w}}{\left | v \right |\left | w \right |}\\=\frac{6-4}{\sqrt(2^2+1^2)\sqrt(3^2+4^2)} =\frac{2}{5\sqrt(5)}[/tex]
velocity of the ball in direction of the the wind
[tex]\left | vcos\theta \right |\\\left | v \right |cos\theta\\\sqrt(2^2+1^2)\frac{2}{5\sqrt(5)} = \frac{2}{5}[/tex]
The size of the velocity of the ball in the direction of the wind is 2/5 ms.
Calculation of the size of velocity:Since there are two vectors v and w
Also, here we assume θ be angle b/w the two vector.
So
Cos θ = 6-4 / √(2^2 + 1^2) √(3^2 + 4^2)
= 2/5√5
Now the velocity of the ball should be
= √(2^2 + 1^2) 2 ÷ 5√(5)
= 2 /5
hence, The size of the velocity of the ball in the direction of the wind is 2/5 ms.
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Light with an intensity of 1 kW/m2 falls normally on a surface with an area of 1 cm2 and is completely absorbed. The force of the radiation on the surface is
Answer:
The force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
Explanation:
Given;
intensity of light, I = 1kw/m² = 1000 W/m²
area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²
Since the light is completely absorbed, the force of the radiation is given by;
F = P/c
where;
c is the speed of light = 3 x 10⁸ m/s
But P = IA
F = IA /c
F = (1000 X 1 X 10⁻⁴) / 3 x 10⁸
F = 3.33 x 10⁻¹⁰ N
Therefore, the force of the radiation on the surface is 3.33 x 10⁻¹⁰ N
The force of radiation will be "3.33 × 10⁻¹⁰ N"
Intensity and ForceAccording to the question,
Intensity of force, I = 1 kW/m² or,
= 1000 W/m²
Area of surface, A = 1 cm² or,
= 1 × 10⁻⁴ m²
Speed of light, c = 3 × 10³ m/s
As we know the relation,
→ F = [tex]\frac{P}{c}[/tex]
or,
P = IA
or,
F = [tex]\frac{IA}{c}[/tex]
By substituting the values, we get
= [tex]\frac{1000\times 1\times 10^{-4}}{3\times 10^3}[/tex]
= 3.33 × 10⁻¹⁰ N
Thus the response above is correct.
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(a) If electrons were used (electron microscope), what minimum kinetic energy would be required for the electrons
Answer:
K = 1.6 10⁻¹⁵ J
Explanation:
In an electron microscope, electrons are used to form images, these electrons are accelerated in electric fields so that they have a kinetic energy that allows obtaining a good amplification with the microscope.
electrical potential energy is converted to kinetic energy
U = K
e V = ½ m v²
v = √2eV /m
the wavelength of these electrons we obtain from the de Broglie equation
λ = h / p
p = mv
λ = h / mv
λ = h / mra 2eV / m
λ = h / ra 2eVm
where we can see that as the potential energy increases, it electrifies the shorter the wavelength of the electrons and consequently the greater the magnification of the microscope
in general these microscopes use from 10000X onwards therefore for this saponification
K = e V
K = 1.6 10⁻¹⁹ 10000
K = 1.6 10⁻¹⁵ J
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
In an inertia balance, a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the action of a set of springs. If a 1.00-kg body vibrates at 1.00 Hz, a 2.00-kg body will vibrate at Group of answer choices
Answer;
a 2.00-kg body will vibrate at 0.707Hz
Answer:-7.9
Explanation:
Kasek rides his bicycle down a 6.0° hill (incline is
6° with the horizontal) at a steady speed of 4.0
m/s. Assuming a total mass of 75 kg (bicycle and
Kasek), what must be Kasek's power output to
climb the same hill at the same speed?
Answer:
P = 2923.89 W
Explanation:
Power is
P = F v
for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane
X Axis
F - Wₓ = 0
F = Wₓ
Y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components of the weight
sin 6 = Wₓ / W
cos 6 = W_{y} / W
Wₓ = W sin 6
W_{y} = W cos 6
F = mg cos 6
F = 75 9.8 cos 6
F = 730.97 N
let's calculate the power
P = F v
P = 730.97 4.0
P = 2923.89 W
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.
Calculate di/dt
di/dt = _________.
Answer:
[tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 450 \ turns[/tex]
The radius is [tex]r = 1.17 \ cm = 0.0117 \ m[/tex]
The position from the center consider is x = 3.45 cm = 0.0345 m
The induced emf is [tex]e = 8.20 *10^{-6} \ V/m[/tex]
Generally according to Gauss law
[tex]\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A[/tex]
Where A is the cross-sectional area of the solenoid which is mathematically represented as
[tex]A = \pi r ^2[/tex]
=> [tex]e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}[/tex]ggl;
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
=> [tex]\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}[/tex]
=> [tex]\frac{di}{dt} = 7.31 \ A/s[/tex]
The value of di/dt from the given values of the solenoid electric field is;
di/dt = 7.415 A/s
We are given;
Number of turns; N = 450 per m
Radius; r = 1.17 cm = 0.0117 m
Electric Field; E = 8.2 × 10⁻⁶ V/m
Position of electric field; r' = 3.45 cm = 0.0345 m
According to Gauss's law of electric field;
∫| E*dl | = |-d∅/dt |
Now, ∅ = BA = μ₀niA
where;
n is number of turns
i is current
A is Area
μ₀ = 4π × 10⁻⁷ H/m
Thus;
E(2πr') = (d/dt)(μ₀niA) (negative sign is gone from the right hand side because we are dealing with magnitude)
Since we are looking for di/dt, then we have;
E(2πr') = (di/dt)(μ₀nA)
Making di/dt the subject of the formula gives;
di/dt = E(2πr')/(μ₀nA)
Plugging in the relevant values gives us;
di/dt = (8.2 × 10⁻⁶ × 2 × π × 0.0345)/(4π × 10⁻⁷ × 450 × π × 0.0117²)
di/dt = 7.415 A/s
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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.
Two identical trucks have mass 5500 kg when empty, and the maximum permissible load for each is 8000 kg. The first truck, carrying a 3900 kg, is at rest. The second truck plows into it at 64 km/h, and the pair moves away at 44 km/h. As an expert witnes, you're asked to determine whether the second truck was overloaded. What do you report? Yes the truck is overloaded, or no, the truck is not overloaded?
Answer:
no, the truck is not overloaded
Explanation:
The computation is shown below;
Let us assume the mass of the loan in the second truck be M
So, the equation is as follows
{(Mass + M) × second truck × 1000 ÷ 3,600} = {(Mass + M + mass + first truck) × Pair moves away × 1,000 ÷ 3,600}
{(5500 + M) × 64 × 1,000 ÷ 3,600 = {(5,500 + M + 5,500 + 3,900) × 44 × 1,000 ÷ 3,600}
(5500 + M) × 64 = (14,900 + M) × 44
352,000 + 64 M = 655,600 + 44 M
After solving this
M = 15,180 kg
Therefore the second truck is not overloaded
A dentist uses a concave mirror (focal length 2 cm) to examine some teeth. If the distance from the object to the mirror is 1 cm, what is the magnification of the tooth
Answer: 2
Explanation:
1/2=1/1 +1/x
x=-2
magnification= 2/1
magnification=2
Air flows through a converging-diverging nozzle/diffuser. A normal shock stands in the diverging section of the nozzle. Assuming isentropic flow, air as an ideal gas, and constant specific heats determine the state at several locations in the system. Solve using equations rather than with the tables.
Answer:
HELLO your question has some missing parts below are the missing parts
note: The specific heat ratio and gas constant for air are given as k=1.4 and R=0.287 kJ/kg-K respectively.
--Given Values--
Inlet Temperature: T1 (K) = 325
Inlet pressure: P1 (kPa) = 560
Inlet Velocity: V1 (m/s) = 97
Throat Area: A (cm^2) = 5.3
Pressure upstream of (before) shock: Px (kPa) = 207.2
Mach number at exit: M = 0.1
Answer: A) match number at inlet = 0.2683
B) stagnation temperature at inlet = 329.68 k
C) stagnation pressure = 588.73 kPa
D) ) Throat temperature = 274.73 k
Explanation:
Determining states at several locations in the system
A) match number at inlet
= V1 / C1 = 97/ 261.427 = 0.2683
C1 = sound velocity at inlet = [tex]\sqrt{K*R*T}[/tex] = [tex]\sqrt{1.4 *0.287*10^3}[/tex] = 361.427 m/s
v1 = inlet velocity = 97
B) stagnation temperature at inlet
= T1 + [tex]\frac{V1 ^2}{2Cp}[/tex] = 325 + [tex]\frac{97^2}{2 * 1.005*10^{-3} }[/tex]
stagnation temperature = 329.68 k
C) stagnation pressure
= [tex]p1 ( 1 + 0.2Ma^2 )^{3.5}[/tex]
Ma = match number at inlet = 0.2683
p1 = inlet pressure = 560
hence stagnation pressure = 588.73 kPa
D) Throat temperature
= [tex]\frac{Th}{T} = \frac{2}{k+1}[/tex]
Th = throat temperature
T = stagnation temp at inlet = 329.68 k
k = 1.4
make Th subject of the relation
Th = 329.68 * (2 / 2.4 ) = 274.73 k
Scientists today learn about the world by _____. 1. using untested hypotheses to revise theories 2. observing, measuring, testing, and explaining their ideas 3. formulating conclusions without testing them 4. changing scientific laws
Answer:
Option 2 (observing, measuring, testing, and explaining their ideas) is the correct choice.
Explanation:
A traditional perception of such a scientist is those of an individual who performs experiments in some kind of a white coat. The reality of the situation is, a researcher can indeed be described as an individual interested in the comprehensive as well as a recorded review of the occurrences occurring in nature but perhaps not severely constrained to physics, chemistry as well as biology alone.The other three choices have no relation to a particular task. So the option given here is just the right one.
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Two ice skaters push off against one another starting from a stationary position. The 45.0-kg skater acquires a speed of 0.375 m/s. What speed does the 60.0-kg skater acquire in m/s
Answer:
0.2812
Explanation:
Given that
mass of skater 1, m1 = 45 kg
mass of skater 2, m2 = 60 kg
speed of skater 1, v1 = 0.375 m/s
To attempt this question, we would be using the Law of conservation of momentum That says the momentum is constant, before and after the movement.
Thus, momentum p = mv
Law of conservation of momentum infers that,
m1v1 = m2v2
Now we proceed to substitute our values into the formula.
45 * 0.375 = 60 * v2
v2 = 16.875 / 60
v2 = 0.2812 m/s
Therefore the speed of the second skater has to be 0.2812 m/s
6)the speed of light is approximately 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth. How many miles away is the star from Earth? Express the answer in scientific notation. Use 365 days in 1 year. The star is nothing miles away from Earth.
Answer:
5.2791264*10¹³
Explanation:
Convert the 9 years to seconds and then multiple it by 186000
The star is 4.62 x 10¹⁶ miles away from Earth.
The speed of light is 186,000 miles per second. It takes light from a particular star approximately 9 years to reach Earth. There are 365 days in 1 year, so it takes 9 x 365 = 3285 days for light from the star to reach Earth.
The distance between the star and Earth is 3285 x 186,000 = 608,810,000 miles. In scientific notation, this is 4.62 x 10¹⁶ miles.
Here is the calculation:
distance = speed * time
distance = 186,000 miles/second * 3285 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute
distance = 608,810,000 miles
distance = 4.62 x 10¹⁶ miles
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Suppose a 1300 kg car is traveling around a circular curve in a road at a constant
9.0 m/sec. If the curve in the road has a radius of 25 m, then what is the
magnitude of the unbalanced force that steers the car out of its natural straight-
line path?
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N[/tex]
So, the force has a magnitude of 4212 N
If two firecrackers produce a sound level of 81 dBdB when fired simultaneously at a certain place, what will be the sound level if only one is exploded?
Answer:
77.96dB
Explanation:
Recall that decibels are a unit of measuring intensity of sound, and depend on the logarithm of the intensity
the intensity, measured in decibels is given by:
I(db)=10log(I/I0)
I is the intensity in MKS units; I0 is the threshold intensity for human hearing (10^-12 W/m^2)
Thus, if the two sounds together have a dB of 81, we know:
81=10log(I/I0)
using the data above, we can find the intensity of the two sounds to be
0.000125 W/m^2
therefore, one firecracker has an intensity half of that, or 0.0000625W/m^2
now use this value to find the dB of one firecracker:
I(dB0=10log(0.0000625/10^-12)=77.96dB
A circuit contains a single 220 pF capacitor hooked across a battery. It is desired to store three times as much energy in a combination of two capacitors by adding a single capacitor to this one.
How would you hook it up?
The capacitor is connected in series to the original capacitor
or
The capacitor is connected in parallel to the original capacitor
I believe its parallel
but now What would its value be?
Answer
The capacitor should be connected in parallel as parallel connection gives the arithmetic sum of capacitance which will give a corresponding sum of energy while capacitors in series gives the sum of the reciprocal if the individual capacitance
Describe how you expect the waveform and the sound you hear changes when you hit the tuning fork harder.
Answer:
In a tuning fork, two basic qualities of sound are considered, they are
1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.
2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.
Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.
Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.
A dipole is oriented along the x axis. The dipole moment is p (= qs). (Assume the center of the dipole is located at the origin with positive charge to the right and negative charge to the left.)
Calculate exactly the potential V (relative to infinity) at a location x, 0, 0 on the x axis and at a location 0, y, 0 on the y axis, by superposition of the individual 1/r contributions to the potential. (Use the following as necessary: q, ε0, x, s and y.)
Answer:
Explanation:
dipole moment = qs = q x s
= charge x charge separation
charge = q
separation between charge = s
half separation l = s / 2
dipole has two charges + q and - q separated by distance s .
Potential at distance x along x axis due to + q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{q}{x-l}[/tex]
Potential at distance x along x axis due to - q
[tex]v_2=\frac{1}{4\pi \epsilon } \times\frac{-q}{x+l}[/tex]
Total potential
v = v₁ + v₂
[tex]v=\frac{1}{4\pi \epsilon } \times( \frac{q}{x-l}-\frac{q}{x+l})[/tex]
[tex]v=\frac{1}{4\pi \epsilon } \times\frac{2ql}{x^2-l^2}[/tex]
[tex]v=\frac{1}{4\pi \epsilon } \times\frac{qs}{x^2-(\frac{s}{2}) ^2}[/tex]
Potential at distance y along y axis due to + q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]
Potential at distance y along y axis due to - q
[tex]v_1=\frac{1}{4\pi \epsilon } \times\frac{-qs}{(y^2+\frac{s^2}{4})^\frac{1}{2} }[/tex]
Total potential
v = v₁ + v₂
[tex]v= 0[/tex]
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any objects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3
Answer:
v = - 14.08 m / s
Explanation:
The definition of center of mass is
[tex]x_{cm}[/tex] = 1 /M ∑sun [tex]x_{i} m_{i}[/tex]
where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.
The velocity of the center of mass can be found by deriving this expression with respect to time
[tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi
let's find the total mass
M = m₁ + m₂ + m₃ + m₄
M = 1.45 + 2.81 +3.89 + 5.03
m = 13.18 kg
let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0
0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)
v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄
let's substitute the given values
v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03
They ask us for the calculations for a time t = 2.83 s
v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03
v = - 14.08 m / s
The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
The given parameters;
[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]
The velocity of the center mass of the particles is calculated as;
[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]
The velocity of particle 1 at time, t = 2.83 s;
[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]
The velocity of particle 2 at time, t = 2.83 s;
[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]
The velocity of particle 3 at time, t = 2.83 s;
[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]
The velocity of the particle 4 at time, t = 2.83 s;
[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]
Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
Learn more here:https://brainly.com/question/22698801
A 50kg block slides down a slope that forms an angle of 54 degrees if it is known that when descending it has a force of 40N and a coefficient of friction of 0.33. What is the acceleration in the block?
Answer:
The acceleration in the block is 2.1 m/s²
Explanation:
Given that,
Mass = 50 kg
Angle = 54°
Force = 40 N
Coefficient of friction = 0.33
We need to calculate the acceleration in the block
Using balance equation
[tex]F_{net}=F_{f}-F\cos\theta[/tex]
[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]
[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]
[tex]a=2.1\ m/s^2[/tex]
Hence, The acceleration in the block is 2.1 m/s²
Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the initial intensity I0, is the intensity of the light that passes through a polarizer if the polarizer is tilted 22.5° from the horizontal?
Answer: I0*0.853
Explanation:
Ok, the Malus's law says that:
If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.
In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°
Then:
I(22.5°) = I0*cos^2(22.5°) = I0*0.853
