Answer:
Speed of light =m/s
wavelength = m
frequency = ?
we have
Speed = frequency × wavelength
[tex]3* 10^8[/tex] = frequency × [tex]5 * 10^{-7}[/tex]
Frequency = [tex]\frac{3*10^8}{5*10^{-7}}=6*10^{14}[/tex]hz
Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Definition of wavelengthFirst of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
Definition of frequencyOn the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Definition of propagation speedFinally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.
Frequency of light waves with wavelength of 5×10⁻⁷ mIn this case, you know:
v= 3×10⁸ m/sf= ?λ= 5×10⁻⁷ mReplacing in the definition of propagation speed:
3×10⁸ m/s = f× 5×10⁻⁷ m
Solving:
3×10⁸ m/s ÷ 5×10⁻⁷ m= f
f= 6×10¹⁴ Hz
In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
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Which of the following statements is NOT true of a hypoosmotic solution?
A. It is also known as a hypotonic solution.
B. There are more dissolved solids within the cell than outside the
cell.
C. It may cause water to move into the cell, which will cause it to
swell.
D. It may cause the cell to shrink or crenate.
Answer:
Good luck on the test, you donuts.
Explanation:
which alkali metal is most reactive
The diagram shows the tropic levels in a lake in a farm. Suppose the farmer sprays a field with a chemical fertilizer. Based on the flow of energy in the lake, which organisms would contain the highest concentration of the chemical fertilizer?
Based on the flow of energy that's in the lake, it can be deduced that the diatoms and algae will have the highest concentration.
Energy flow refers to the flow of energy through living organisms that are within an ecosystem. Energy is passed from one trophic level to the next trophic level.
If the farmer sprays a field with chemical fertilizer, the organism that would contain the highest concentration of the chemical fertilizer are diatoms and algae since they're at the bottom of the ecosystem.
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A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angular velocity ω of the wheel at the bottom of the incline?
Express your answer in radians per second.
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]
[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]
[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]
[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]
[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]
[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]
[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
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The angular velocity of the wheel depends on the mass, radius and the
mode of rotation of the wheel (with or without slipping).
The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/secReasons:
The given parameters are;
Radius of the wheel, r = 2.0 m
Height of the incline, h = 8.0 m
Required:
Angular velocity of the wheel at the bottom of the incline.
Solution:
The potential energy of the wheel at the top of the hill, P.E. = m·g·h
[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]
Where;
v = The translational velocity of the wheel = ω·r
I = The moment of inertia of the wheel = m·r²
Therefore'
[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]
At the bottom of the hill, the potential energy is converted to kinetic energy
Therefore;
P.E. = Sum of K.E.
m·g·h = m·r²·ω²
g·h = r²·ω²
[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]
The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/secLearn more about the law of conservation of energy here:
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if a current of 2.0 A is flowing from point a to point b, the potential difference between Vb- Va (in V) is:
a. 6
b. 8
c. -6
d. -8
e. 22
Answer:
[tex]\huge\color{skyblue}\boxed{\colorbox{black}{Answer ☘}}[/tex]
[tex]total \: resistance \: ( R_{t} ) = (3 + 1)ohm \\ (since \: the \: resistors \: are \: connected \\ in \: series)[/tex]
[tex]current \: flowing \: through \: circuit(I) = 2A \\ \\ now... \: by \: ohms \: law \\ V = IR\\ V = (2)(4) \\ = > 8v[/tex]
therefore , option (b) is correct!!hope helpful~
When a 1 is on the input of an inverter, what is the output?
Answer:
When the input to an inverter is high (1) the output is low (0); and when the input is low, the output is high.
PLEASE HELP ASAP
I need help understanding this question, form my guess I’m thinking it’s C but I’m not to sure
Answer:
Explanation:
Without the diagram, we have to bracket our answers.
Total energy is constant, so we eliminate options A and B.
C is correct if position 1 is closer to the star than position 2
D is correct if position 2 is closer to the star than position 1
A person is drinking a glass of soda with ice.
which option describes the relative kinetic energy of molecules in and above the soda in the glass?
A. in : least energy
above : intermediate and greatest energy
B. in : greatest energy
above : least energy
C. in : least energy
above : greatest energy
D. in least and intermediate energy
above : greatest energy
The relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.
The relative kinetic energy of gas molecules increases with increase in the mean distance between the gas molecules.
Also, relative kinetic energy of gas molecules increases with in the temperature of the gas molecules and decreases with a decrease in the temperature of the of the gas molecules;
ΔK.E ∝ T
The ice in the soda lowers the temperature of the gas molecules, thereby reducing their average speed which in turn reduces the average kinetic energy of the gas molecules in the soda.
Above the soda in the glass, the concentration of the gas molecules is less and their mean distance is greatest when compared to inside the soda. This results to an increase in the speed of the gas molecules which increases their average kinetic energy.
Thus, the relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.
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HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.
The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
The given parameters;
Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, [tex]\lambda_{rms}[/tex] = 4.32 x 10⁻⁸ mMean collision time, [tex]\tau = 3 \times 10^{-10} \ s[/tex]The mean velocity of the gas molecules is calculated as follows;
[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]
The temperature of the gas molecules is calculated as follows;
[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]
where;
k is Boltzmann constant[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]
The number of gas molecules per unit volume is calculated as follows;
[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]
The pressure of the gas molecule is calculated as follows;
[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]
Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
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8/10
A person stands on a square pedestal whose side lengths are
1.20 m. The person and pedestal together weigh 1.84 x 103 N.
What is the pressure the person and and pedestal exert on the
floor in Pa?
Answer:
Explanation:
P = 1.84 x 10³ N / 1.20² m² = 1,277.77777... N/m²
P = 1.28 x 10³ Pa
a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed of the car is
Answer:
The new speed of the car is 30 m/s.
Hope you could get an idea from here.
Doubt clarification - use comment section.
A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s
The speed of the cannon after firing is 1.2 m/s
This can be solved using the law of conservation of momentum.
From the law of conservation of momentum,
⇒ The momentum of the projectile is equal to the momentum of the cannon.
MV = P................ Equation 1⇒ Where :
M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.⇒ make V the subject of the equation
V = P/M.................. Equation 2From the question,
⇒ Given:
P = 600 kgm/sM = 500 kg⇒ Substitute these values into equation 2
V = 600/500V = 1.2 m/sHence, The speed of the cannon after firing is 1.2 m/s
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If a rocket experiences an acceleration generated by the gravity force between the earth and itself, what is this acceleration if the rocket flies 1000 km above the ground and the Earth's radius is 6.378 * 10 ^ 6 * r m. We know the Earth has a mass of 5.97*10^ 24 kg(in m/s^ 2 , G=6.67*10^ -11 N(m/kg)^ 2 ) ?
a 8.97
b 7.32
c 9.81
d 5.5
e 11.45
This question involves the concepts of Gravitational Force and Weight force.
The value of acceleration is "b. 7.32 m/s²".
At the given height the weight of the rocket must be equal to the gravitational force between rocket and the Earth:
[tex]W=F_G\\mg=\frac{GmM}{R^2}\\\\g=\frac{GM}{R^2}[/tex]
where,
g = acceleration = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = mass of earth = 5.97 x 10²⁴ kg
R = Radius of Earth + Height = 6.378 x 10⁶ m + 1 x 10⁶ m = 7.378 x 10⁶ m
Therefore,
[tex]g=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{(7.378\ x\ 10^6\ m)^2}[/tex]
g = 7.32 m/s²
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find the total volume of the material used to making the cylinder
Answer:
Hey mate
Explanation:
Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.
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is xenon a pure substance
[tex]\large\huge\green{\sf{Yes}}[/tex]
Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.
Answer:
Explanation:
vA / vB = √(TA/(m/L)) / √(TB/(m/L))
The lengths are the same, so the L divides out to 1
The material is identical so the mass will be directly proportional to the cross sectional area of the string
vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))
π/4 is common so divides out to 1
vA / vB = √(TA/dA²) / √(TB/dB²)
vA / vB = √(410/0.489²) / √(809/1.27²)
vA / vB = 41.407 / 23.396
vA / vB = 1.8488961...
vA / vB = 1.85
2. Where do hyperbolic comets originate?
A.the Oort cloud
B.the asteroid belt
B.the Kuiper belt
C.interstellar space
Answer:
Hyperbolic comets originate from the Kuiper Belt.
In the following free body diagram, what is the net force on the object?
A.10n
B.5 N to the right
C.20 N to the right
D.7 N to the right
Answer:
B
Explanation:
Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.
ΣF = 7 + 18 + (-20) = 5N to the right
Raghu studies in grade 6th. He wants a cricket bat to be made by a carpenter. He tells the
that the length of the bat should be 7 hand spans. The tall carpenter tells Raghu that it
will be ready by tomorrow. When Raghu went to collect the bat the next day, he was very
disappointed. Why? Was the bat longer or shorter than what Raghu expected? Give reason.
carpenter
Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.
From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.
Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.
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Convert the decimal number 61078 to binary by using sum-of-weights method
Answer:
1110111010010110
Explanation:
I am not able to upload the working out using the sum of weights method sorry
a car moving at 5 m/s accelerates at a rate of 10 m/s2 for 25 seconds. How far does it move during this time?
Answer:
t is time in s For example, a car accelerates in 5 s from 25 m/s to 3 5m/s. Its velocity changes by 35 - 25 = 10 m/s. Therefore its acceleration is 10 ÷ 5 = 2 m/s2
Explanation:
E
D
С
B
A
Column
Column
1.
Electron Cloud
B
2
Nucleus
b. A
3.
Electron
CD
4.
Neutron
d.
5.
Proton
ec
Answer:
1. D electron cloud
2. A nucleus
3. E electron
4. B neutron
5. C proton
Explanation:
electrons are located outside the nucleus. The nucleus holds the protons and neutrons. Protons have a positive symbol (+). electron cloud holds the electrons.
Explanation:
Electron cloud : D
Nucleus : A
Electron : E
Neutron : B
Proton : C
hope this helps you.
A ball has the energy to move 30 m/s with the mass of 5. What is the energy
What happens to the gravitational force between two objects if the distance between them triples?
A. The force increases by a factor of 9
B. The force decreases by a factor of 9
C. The force decreases by a factor of 3
D. The force increases by a factor of 3
Hello I am absolutely stumped on these six physics problems. Please help me on them.
Answer:
Explanation:
20° from the normal = 110° from parallel
1a) τ = (200sin90)[6] + (75sin110)[3] - (100sin110)[3]
τ = 1,129.52305... = 1100 N•m CCW
1b) τ = 200(6)(sin90) + 75(-3)(sin(360-110)) + 100(3)(sin(270 + 20)
τ = 1,129.52305... = 1100 N•m CCW
1c) My directions agree, both are positive z by right hand rule.
1d) Moment of inertia for a thin rod about an axis perpendicular to its center is
I = (1/12)mL²
τ = Iα
α = τ/I = 1129.523 / ((1/12)(200)(12²)) = 0.4706 rad/s²
θ = ½αt²
θ = ½(0.4706)(2•60)² = 3,388.56916... radians
θ = 3400 radians
at which time is is spinning about 9 revolutions per second
Water pumps use electrical energy to create mechanical energy needed to move water.
A 1875 W water pump is plugged into a 110 V electrical outlet.
Answer:
n/a
Explanation:
this question is not feasible
What effect does the time taken to lift the mass have on power output?
You are pulling a sled with a constant velocity using a rope held horizontally in
the snow. Using a spring scale attached to the rope, you measure the tension
to be 0.96 N. The sled's mass is 0.58 kg. Write the surface force the ground
exerts on the sled in component form, and determine its magnitude.
Answer:
Explanation:
if you are pulling in the positive direction and Upward is also positive.
F = -0.96i + 0.58(9.8)j
F = -0.96i + 5.7j
F = √(-0.96² + 5.7²) = 5.8 N
You are pulling a sled with constant velocity by using a rope then the surface force the ground exerts on the sled will be 5.8 N.
What is tension?The pulling force conveyed axially by a string, cable, chain, or another analogous object, or by either end of a rod, truss member, or another such three-dimensional object, is referred to as tension in physics. Another way to think of tension is as the action-reaction pair of forces acting at each end of the previous elements. Perhaps tension is the polar opposite of compression.
A restoring force may cause what is also referred to as tension when atoms or molecules are torn apart from one another at the atomic level and build up potential energy.
If you are moving upward and pushing in a positive way,
F = -0.96 i + 0.58(9.8) j
F = -0.96 i + 5.7 j
F = [tex]\sqrt{(-0.96^2 + 5.7^2)}[/tex]
F= 5.8 N
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A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth of the bottom end of the cylinder
Answer:
Explanation:
I will ASSUME that the cylinder floats with the axis through the center of its flat ends is perpendicular to the water surface.
The cylinder will displace it's mass in water.
water density has a value of 1000 kg/m³
if the depth of the bottom is "h" in meters
1.80 kg = π(0.018² m²)(h m)(1000) kg/m³
h = 1.768388256...
or about 1.77 m
The following is an example of what type of reaction?
Answer:
Single Displacement Reaction
Explanation:
Here, only one element is getting replaced during the reaction, that is, silver gets replaced by copper, hence, single displacement reaction.
Hope it helps :)