Answer:
The answer is magma
Explanation:
I'm 99.99% this is right
Answer: its magnetic reversal
Explanation: jus took the test make brainly
Why does a solid keep its shape
Answer:
Solids can hold their shape because their molecules are tightly packed together. Atoms and molecules in liquids and gases are bouncing and floating around, free to move where they want. The molecules in a solid are stuck in a specific structure or arrangement of atoms.
Please help I would appreciate it
Answer:
3.176hours
Explanation:
270/85=3.126 hours DISTANCE / SPEED = TIME
The table shows information about two of the major moons of Uranus.
Moon Orbital Period (days) Average Distance from Uranus (km)
Miranda 0.319
129,390
Titania
8.71
435,910
What is the average distance between Miranda and Uranus?
2.84 104 km
1.29 x 105 km
1.47 x 106 km
4.65 x 107 km
Answer:
B. 1.29 x 10^5
Answer:
B. 1.29 x 10^5
Explanation:
just took the test on ed2020
Rank the following cubes in order of the amount of electric flux through their surfaces, from most positive to?
Rank the following cubes in order of the amount of electric flux through their surfaces, from most positive to most negative.
A cube with sides 10 cm long that contains a + 2.00 micro-coulomb point charge
A cube with sides 10 cm long that contains a + 1.00 micro-coulomb point charge
A cube with sides 20 cm long that contains a + 2.00 micro-coulomb point charge
A cube with sides 20 cm long that contains a + 1.00 micro-coulomb point charge
Answer: Ranking : a = c > b > d
Explanation:
The amount of electric flux is directly proportional to the amount of charge enclosed.
The greater the charge enclosed, the greater the electric flux through their surface
a)
A cube with sides 10 cm long that contains a + 2.00 micro-coulomb point charge
∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C
b)
A cube with sides 10 cm long that contains a + 1.00 micro-coulomb point charge
∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C
c)
A cube with sides 20 cm long that contains a + 2.00 micro-coulomb point charge
∅ = Q/∈₀ = (2.00 x 10⁻³) / (8.85 x 10⁻¹²) = 2.26 x 10⁸ Nm²/C
d) A cube with sides 20 cm long that contains a + 1.00 micro-coulomb point charge
∅ = Q/∈₀ = (1.00 x 10⁻³) / (8.85 x 10⁻¹²) = 1.13 x 10⁸ Nm²/C
Ranking :
a = c > b > d
Answer:
Using Gauss's Law, the total amount of electric flux through a closed surface is proportional to the charge enclosed. (The surface integral of the electric field over the closed surface equals the enclosed charge divided by the constant of proportionality, the permittivity of free space.) So, since surfaces 1 and 3 have the most positive enclosed charge, they have the most positive electric flux (they have the same). Since surfaces 2 and 4 have the least positive enclosed charge, they have the least positive electric flux. and are the same
thus LHS to RHS; a=c > b=d
What is the displacement of the object?
Answer:
Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Which of the following best decribes the velocity of an object?
a
30 m/s
b
30 m east
С
30 m/s east
d
30 m
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volume charge density rho(r) of this thin shell distribution in terms of σ and an appropriate delta function. Verify explicitly that the units of your final expression are correct. Also show that your total integrated charge comes out right.
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴
[tex]\rho (r) \ \alpha \ \delta (r -R)[/tex]
[tex]\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)[/tex]
[tex]\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)[/tex]
To find the constant k, we examine the total charge Q which is:
[tex]Q = \int \rho (r) \ dV = \int \sigma \times dA[/tex]
[tex]Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2[/tex]
∴
[tex]\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2[/tex]
[tex]\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]
[tex](2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2[/tex]
Thus;
[tex]k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2[/tex]
[tex]k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2[/tex]
[tex]k * R^2= \sigma \times R^2[/tex]
[tex]k = R^2 --- (2)[/tex]
Hence, from equation (1), if k = [tex]\sigma[/tex]
[tex]\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}[/tex]
[tex]\mathbf{\rho (r) =0 \ \ at \ \ r<R \ \ or \ \ r>R}[/tex]
To verify the units:
[tex]\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}[/tex]
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
[tex]Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr[/tex]
[tex]Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr[/tex]
[tex]Q = 4 \pi \sigma *R^2[/tex] since [tex]( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )[/tex]
[tex]\mathbf{Q = 4 \pi R^2 \sigma }[/tex]
Una cubeta de agua puede girar en un círculo vertical sin que el agua se derrame, incluso en lo alto del círculo cuando la cubeta está boca abajo. ¿Por qué sucede esto? ¿Que otros ejemplos puedes mencionar que utilicen el mismo principio?
Answer:
v> √ gr
Explanation:
Let's analyze this problem from Newton's second law
At the lowest point
N - W = m a
at the highest point
W = m a’
in both cases there is a net force towards the center of the circle that we can call the centripetal force, which is responsible for changing the direction and magnitude of the acceleration.
When the bucket is in the highest part, the centripetal force is equal to the weight of the water, but since it carries a horizontal speed, until it starts to fall, it is moving and therefore they follow the bottom of the tube. This implies that there is a minimum speed for this to occur
v> √ gr
This principle is applied in many things, for example the roller coaster, centrifuges, simulators of the effect of acceleration on people.
Some common types of forces that you will be dealing with include the gravitational force (weight), the force of tension, the force of friction, and the normal force. It is sometimes convenient to classify forces as either contact forces between two objects that are touching or as long-range forces between two objects that are some distance apart. Contact forces include tension, friction, and the normal force. Long-range forces include gravity and electromagnetic forces. Note that such a distinction is useful but not really fundamental: For instance, on a microscopic scale the force of friction is really an electromagnetic force. In this problem, you will identify the types of forces acting on objects in various situations.
Now consider a different situation. A string is attached to a heavy block and is used to pull the block to the right along a rough horizontal table.
a. What is the upward force that acts on the book called?
1-tension
2-normal force
3-weight
4-friction
b. Which object exerts a force on the block that is directed to the right?
1-the block itself
2-the earth
3-the surface of the table
4-the string
Answer:
No. A:
1-tension
No. B:
4-the string
Explanation:
The upward force that acts on the book is called "tension" while the "string" is the object that exerts a force on the block that is directed to the right.
As the string is pulled, the tension exerts an upward force on the block. The frictional force acts on the block to the left. So, both the tension and friction will act on the block in order to effect its pulling on the surface of the table.
One end of a spring with a force constant of k = 10.0 N/m is attached to the end of a long horizontal frictionless track and the other end is attached to a mass m = 2.20 kg which glides along the track. After you establish the equilibrium position of the mass-spring system, you move the mass in the negative direction (to the left), compressing the spring 2.48 m. You then release the mass from rest and start your stopwatch, that is x(t = 0) = −A, and the mass executes simple harmonic motion about the equilibrium position. Determine the following.(a) displacement of the mass (magnitude and direction) 1.0s after it is released
(b) velocity of the mass (magnitude and direction) 1.0s after it is released
(c) acceleration of the mass (magnitude and direction) 1.0s after it is released
(d) force the spring exerts on the mass (magnitude and direction) 1.0s after it is released
(e) How many times does the object oscillate in 12.0s?
Answer:
-2.478
0.379
11.14
24.78
Explanation:
Angular frequency of spring in harmonic motion is given by?
ω = √(k/m)
ω = √(10/2.2)
ω = √4.54
ω = 2.13 s^-1
If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine
x(t) = -A * cos(ωt)
x(t) = -2.48 * cos(2.13 * 1)
x(t) = -2.48 * 0.9993
x(t) = -2.478
Velocity and acceleration are 1st and 2nd derivative of position
b)
v(t) = Aω * sin(ωt)
v(t) = 2.48 * 2.13 * sin(2.13 * 1)
v(t) = 5.282 * sin2.13
v(t) = 5.282 * 0.03717
v(t) = 0.379 m/s
c)
a(t) = Aω^2 * cos(ωt)
a(t) = 2.48 * 2.12² * cos(2.13 * 1)
a(t) = 2.48 * 4.494 * cos2.13
a(t) = 11.15 * 0.9993
a(t) = 11.14 m/s²
d)
F = -k * x(t)
F = -10 * -2.478
F = 24.78 N
The heat flux, q, from a hot body in a stationary fluid is known to be a function of the fluid’s thermal conductivity k, and kinematic viscosity, ν, the temperature difference ΔT , the length of the object L, and the product of the local gravitational constant and the thermal expansion coefficient for the fluid, gβ. Determine the number of Pi groups thatcan be formed from these six parameters.
Answer:
The number of pi groups are P = 3 pi groups
Explanation:
From the question we are told that
The number of parameters is n = 6
These parameters are : Fluid thermal conductivity : k
Kinematic viscosity: v
The temperature difference ΔT
The length of the object L
Gravitational constant g
Thermal expansion coefficient β
The number of the basic dimensional unit is N = 3
Mass M
Length L
Time T
Generally the number of Pi groups is mathematically evaluated as
P = n - N
=> P = 6 - 3
=> P = 3 pi groups
The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.
Required:
a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?
b. How do you know the answer in part (B) to be correct?
Complete Question
The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.
Required:
a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?
b
Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.10 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?
c. How do you know the answer in part (B) to be correct?
Answer:
a
The two possible frequencies of the second propeller in Hz
[tex]f_1 = 11.58\ Hz [/tex]
[tex]f_2 = 7.58 \ Hz [/tex]
The two possible frequencies of the second propeller in rpm
[tex]f_1 = 695 \ rpm [/tex]
[tex]f_2 = 455 \ rpm [/tex]
b
The correct answer for the frequency of the second single-bladed propeller is
[tex]f_1 = 695 \ rpm [/tex]
c
The above answer is correct because when the beat frequency of the second propeller increases(i.e from 2.0 Hz to 2.10 Hz) the frequency of the second propeller becomes much greater than that of the first propeller so looking at the two possible value of frequency of the second propeller (i.e 695 rpm and 455 rpm ) we see that it is 695 rpm that is showing that increase of the second propeller compared to the first propeller
Explanation:
From the question we are told that
The number of engines is n = 2
The number of propellers is m = 2
The angular frequency of the single-bladed propellor [tex]w = 575 rpm[/tex]
The frequency of the beat heard at this velocity is [tex]f = 2.0 \ Hz[/tex]
Converting the beat frequency to rpm
[tex]f = 2 * 60 = 120 \ rpm[/tex]
Generally the the two possible frequencies of the second propeller in rpm is
[tex]f_1 = w + f[/tex]
=> [tex]f_1 = 575 + 120[/tex]
=> [tex]f_1 = 695 \ rpm [/tex]
And
[tex]f_2 = w - f[/tex]
=> [tex]f_2 = 575 - 120[/tex]
=> [tex]f_2 = 455 \ rpm [/tex]
Converting the angular frequency of the single-bladed propellor to rpm
[tex]w = \frac{575}{60} [/tex]
[tex]w = 9.58 \ Hz [/tex]
Generally the the two possible frequencies of the second propeller in
Hz is
[tex]f_1 = w + f[/tex]
=> [tex]f_1 = 9.58 + 2[/tex]
=> [tex]f_1 = 11.58\ Hz [/tex]
And
[tex]f_2 = w - f[/tex]
=> [tex]f_2 = 9.58 - 2[/tex]
=> [tex]f_2 = 7.58 \ Hz [/tex]
giving brainlist down below pick one form of government that is easy to do
Direct Democracy, Representative Democracy, Dictatorship, Oligarchy, Communism, or Socialism
Answer:
Representative Democracy.
Explanation:
It is simple and easy because you choose a representative to make choices for the good of your people. It is much simpler then all the rest.
A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels
640 m in 40 s. For the 40 s interval, find (a) the average velocity, (b) the final velocity, and (c) the
acceleration.
Answer:
(a) The average velocity is 16 m/s
(b) The acceleration is 0.4 m/s^2
(c) The final velocity is 24 m/s
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:
[tex]v_f=v_o+at\qquad\qquad [1][/tex]
The distance traveled by the object is given by:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]
(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.
We know the distance is x=640 m and the time taken t= 40 s, thus:
[tex]\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16[/tex]
The average velocity is 16 m/s
Using the equation [1] we can solve for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
(c) From [2] we can solve for a:
[tex]\displaystyle a= 2\frac{x-v_ot}{t^2}[/tex]
Since vo=8 m/s, x=640 m, t=40 s:
[tex]\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4[/tex]
The acceleration is 0.4 m/s^2
(b) The final velocity is calculated by [1]:
[tex]v_f=8+0.4\cdot 40[/tex]
[tex]v_f=8+16=24[/tex]
The final velocity is 24 m/s
According to the graph given, what is the best explanation for what is occurring as energy us removed?
A) an exothermic change in which water freezes
B) an exothermic change in which water boils
C) and endothermic change in which rubbing alcohol evaporates
D) and endothermic change in which water freezes
Answer:
I believe it's A) "An exothermic change in which water freezes I could be wrong though just giving my opinion
Explanation:
A sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
a. What is the sprinters velocity at the end of the 200 m?
b. How long does it take him to cover it?
c. What is his average velocity?
Answer:
a. 28.28
B.14.4
C.14.4
Explanation:
A. v^2=u^ + 2as
v^2=0^2 + 2*2*200
B. v=u+at
t=v-u/a
C. V+u/2
The sprinter velocity at the end of the 200 m would be 28.28 m/s.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
As given in the problem a sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
By using the third equation of the motion,
v² - u² = 2×a×s
v² - 0² = 2×2×200
v² = 800
v =28.28 m/s
The sprinter velocity at the end of the 200 m would be 28.28 m/s
By using the second equation of motion
S = ut + 1/2*a*t²
u= 0 m/s , a= m/s² s = 200 m
200 = 0 + 0.5*2*t²
t² = 200
t = √200
t = 14.14 seconds
It would take him 14.14 seconds to cover.
average velocity = initial velocity + final velocity /2
= 0+ 28.28/2 m/s
= 14.14 m/s
Thus, The sprinter velocity at the end of the 200 m would be 28.28 m/s.
Learn more about equations of motion from here
brainly.com/question/5955789
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Ancient gold coins have been found in Greece and Italy.What property allow for gold to be stamped into coins ?
Answer: Malleability
Explanation:
Gold in addition to being so rare, is the most malleable element there is. A Malleable element can be shaped and reshaped without it breaking and gold is so malleable that you can hit it till it becomes so flat, its transparent.
This malleability allowed for the ancient peoples to be able to shape and stamp gold into coins and their rarity made them quite valuable. Even today, investors still shape gold into exquisite designs of coins so it not only has monetary value, but aesthetic value as well.
Describe how cool and warm air move in the atmosphere.
Answer:
it is cool and warm
Explanation:
A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.277 m/s^2 in the x direction. After 46.6 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find:
a. The magnitude.
b. The direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answer:
a) 25.76 m/s
b) 30°
Explanation:
See attachment
A change in velocity over time is known as?
Answer:
Acceleration
Explanation:
I am writing this because it must be 20 characters to submit
Which describes the greatest displacement?
A. walking 3 m east, then 3 m north,
then 3 m west
B. walking 3 m east, then 3 m south,
then 3 m east
C. walking 3 m north, then 3 m south,
then 3 m north
D. walking 3 m north, then 3 m west,
then 3 m south
Tabial 1
Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved along a line parallel to the line joining the speakers and 9.4 m from it. An intensity maximum is measured a point P0 where the microphone is equidistant from the two speakers. As we move the microphone away from P0 to one side, we find intensity minima and maxima alternately. Take the speed of sound in air to be 344 m/s, and you can assume that the slits are close enough together that the equations that describe the interference pattern of light passing through two slits can be applied here.
Required:
a. What is the distance, in meters, between Po and the first intensity minimum?
b. What is the distance, in meters, between Po and the first intensity maximum?
c. What is the distance, in meters, between Po and the second intensity minimum?
d. What is the distance, in meters, between Po and the second intensity maximum?
Answer:
a. approximately [tex]1.1\; \rm m[/tex] (first minimum.)
b. approximately [tex]2.2\; \rm m[/tex] (first maximum.)
c. approximately [tex]3.4\; \rm m[/tex] (second minimum.)
d. approximately [tex]4.7\; \rm m[/tex] (second maximum.)
Explanation:
Let [tex]d[/tex] represent the separation between the two speakers (the two "slits" based on the assumptions.)
Let [tex]\theta[/tex] represent the angle between:
the line joining the microphone and the center of the two speakers, andthe line that goes through the center of the two speakers that is also normal to the line joining the two speakers.The distance between the microphone and point [tex]P_0[/tex] would thus be [tex]9.4\, \tan(\theta)[/tex] meters.
Based on the assumptions and the equation from Young's double-slit experiment:
[tex]\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}[/tex].
Hence:
[tex]\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)[/tex].
The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let [tex]\lambda[/tex] denote the wavelength of this wave.
[tex]\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}[/tex].
Calculate the wavelength of this wave based on its frequency and its velocity:
[tex]\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m[/tex].
Calculate [tex]\theta[/tex] for each of these path differences:
[tex]\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}[/tex].
In each of these case, the distance between the microphone and [tex]P_0[/tex] would be [tex]9.4\, \tan(\theta)[/tex]. Therefore:
At the first minimum, the distance from [tex]P_0[/tex] is approximately [tex]1.1\; \rm m[/tex].At the first maximum, the distance from [tex]P_0[/tex] is approximately [tex]2.2\; \rm m[/tex].At the second minimum, the distance from [tex]P_0[/tex] is approximately [tex]3.4\; \rm m[/tex].At the second maximum, the distance from [tex]P_0[/tex] is approximately [tex]4.7\;\rm m[/tex].Interference is the result when two or more waves combine
The distance between P₀ and
a. The first intensity minimum is approximately 1.06 m
b. The first intensity maximum is approximately 2.165 m
c. The second intensity minimum is approximately 3.36 m
d. The second intensity minimum is approximately 4.72 m
The reasons the above values are correct are given as follows:
The known parameters are;
The distance between the two speakers = 0.94 m
Frequency of the tone produced by the two speakers = 1,630 Hz (in sync)
The line along which the microphone moves is parallel to the line between the two speakers
The distance between the parallel lines above = 9.4 m
The speed of sound in air, v₀ = 344 m/s
The interference pattern of light passing between two slits is to be applied
a. Based on the arrangement, we have;
P₀ = 9.4 × tan(θ)
Where;
θ = The angle formed formed by the line from the microphone to the midpoint of the distance between the two speakers and the perpendicular bisector to the line joining the two speakers
Based on Young's double-slit experiment, we have;
[tex]sin(\theta) = \dfrac{Path \ difference}{d}[/tex]
Where;
d = The distance between the two speakers representing the slits
The path difference for a minimum is n × λ/2
Where n = 1, 3, 5,...,(set of odd numbers)
The path difference for a maximum intensity sound is n·λ
Where n = 1, 2. 3, ..., n (n is an integer)
The wavelength, is given as follows;
[tex]\lambda = \dfrac{v}{f}[/tex]
Therefore;
[tex]\lambda = \dfrac{344}{1,630} = \dfrac{172}{815} \approx 0.211[/tex]
The wavelength, λ ≈ 0.211
Therefore, the angle, θ, for the first minima, θ, ≈arcsine(0.211/(2×0.94))
First minima, λ/2, P₀ =9.4 × tan(arcsine(0.211/(2×0.94))) ≈ 1.06
First maxima, λ, P₀ =9.4 × tan(arcsine(0.211/(94))) ≈ 2.165
Second minima, 3·λ/2, P₀ = 9.4 × tan(arcsine(3*0.211/(2×0.94))) ≈ 3.36
Second maxima, 2·λ, P₀ = 9.4 × tan(arcsine(2*0.211/(0.94))) ≈ 4.72
Therefore;
a. The distance between P₀ and the first intensity minimum is ≈ 1.06 m
b. The distance between P₀ and the first intensity maximum is ≈ 2.165 m
c. The distance between P₀ and the second intensity minimum is ≈ 3.36 m
d. The distance between P₀ and the second intensity minimum is ≈ 4.72 m
Learn more about interference of sound waves here:
https://brainly.com/question/14006922
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Think about a hot air balloon travelling around the world in 11 days. How can a balloon travel so far and fast without a engine or other system on board to move the balloon horizontally
Answer:
?
Explanation:
HELP PLEASE THANKS!! Explain why Gravitational forces are always attractive.
Answer:
cause without gravity, the earth will start to move away from the orbit and crash into the sun like a raining meteor of babies diaper falling on the ground of smelly dunken doughnuts
Explanation:
lol
___ is the pull that all objects exert on each other.
A.Resistance
B.Inertia
C.Gravity
D. Force
Answer:
B
Explanation: Inertia is where it exerts.
If the strength of the action force is halved, what happens to the strength of the reaction force?
Answer:
it is reduced to half of that strength
Upon impact, bicycle helmets compress, thus lowering the potentially dangerous acceleration experienced by the head. A new kind of helmet uses an airbag that deploys from a pouch worn around the rider's neck. In tests, a headform wearing the inflated airbag is dropped from rest onto a rigid platform; the speed just before impact is 6.00 m/s. Upon impact, the bag compresses its full 12.0 thickness, slowing the headform to rest.
Required:
Determine the acceleration of the head during this event, assuming it moves the entire 12.0 cm.
Answer:
acceleration = -15.3g
Explanation:
given data
speed = 6.00 m/s.
thickness = 12
moves the entire = 12.0 cm
solution
we will use here equation that is
v² - u² = 2 × a × s ........................1
here v = 0 is the final velocity and u = 6.0 m/s is initial velocity and s= 0.12 m is the distance covered and a is the acceleration
so we put here value and get acceleration
a = [tex]\frac{v^2-u^2}{2s}[/tex]
a = [tex]\frac{0^2-6^2}{2\times 0.12}[/tex]
a = -150 m/s² ( negative sign means it is a deceleration )
and
acceleration in units of g
a = [tex]\frac{-150}{9.8}[/tex]
a = -15.3 g
Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
A) Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two speeds reach their separations?A) Electrons A is moving faster than electron B.B) Electron B is moving faster than electron A.C) Both electrons are moving at the same (nonzero) speed in the opposite direction.D) Both electrons are moving at the same (nonzero) speed in the same direction.E) Both electrons are momentarily stationary.
2) What is the minimum separation rmin that the electrons reach?
Complete Question
Two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
A) Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two speeds reach their separations?
A) Electrons A is moving faster than electron B.
B) Electron B is moving faster than electron A.
C) Both electrons are moving at the same (nonzero) speed in the opposite direction
.D) Both electrons are moving at the same (nonzero) speed in the same direction.
E) Both electrons are momentarily stationary.
2) What is the minimum separation[tex]r_{min}[/tex] that the electrons reach?
Answer:
1
The correct option is E
2
[tex]r_{min} = \frac{kq^2}{4 mv^2}[/tex]
Explanation:
From the question we are told that
The mass of each electron is m
The charge of each electron is q
The speed of electron A is v
The speed of electron B is 3v
Generally at their point of separation the repulsion force is equal to the force that is propelling the electrons due to this the electrons are momentarily stationary
Generally the total initial kinetic energy of both electron is mathematically represented as
[tex]K_{inT} = K_A + K_B[/tex]
=> [tex]K_{inT} = \frac{1}{2}m (v)^2 + \frac{1}{2} m (3v)^2[/tex]
=> [tex]K_{inT} = \frac{1}{2} (mv^2 + 9v^2m)[/tex]
=> [tex]K_{inT} = 5mv^2 [/tex]
Generally the total final kinetic energy of both electron is mathematically represented as
[tex]K_{fT} = \frac{1}{2} *m * v^2 + \frac{1}{2} *m * v^2[/tex]
Here v is the velocity due to the repulsion force
[tex]K_{fT} = mv^2 [/tex]
Generally the final potential energy of the both electrons is
[tex]P_f = \frac{ k * q^2}{r_{min}}[/tex]
Here k is the coulombs constant
So according to energy conservation law
[tex]K_{inT} = K_{fT} + P_f[/tex]
=> [tex]5mv^2 = mv^2 + \frac{ k * q^2}{r_{min}} [/tex]
=> [tex]r_{min} = \frac{kq^2}{4 mv^2}[/tex]
ou step off the limb of a tree clinging to a 30-m-long vine that is attached to another limb at the same height and 30-m distant. Assuming air resistance is negligible, how fast are you gaining speed at the instant the vine makes an angle of 25 degrees with the vertical during your descent
Answer:
v = 7.42 m / s
Explanation:
For this exercise we will use the conservation of mechanical energy.
Starting point. Before jumping from the tree
Em₀ = U = m g h
Final point . Part of the trajectory at 25º
Em_{f} = K + U = ½ m v2 + m g y
as they indicate that there is no air resistance, mechanical energy is conserved
Em₀ = Em_{f}
m g h = ½ m v² + m g y
v² = 2g (h - y)
Let's use trigonometry to find the height that has descended, how the angle is measured with respect to the vertical
cos 25 = y / L
y = L cos 25
we substitute
v² = 2 g (h - L cos 25)
for this case h = L = 30 m
v2 = 2g L (1- cos25)
let's calculate
v² = 2 9.8 30 (1 -cos 25)
v = √55.09
v = 7.42 m / s
An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package.
(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units
Answer:
a
[tex]D = 1162.7 \ m [/tex]
b
[tex]\beta =- 65.55^o[/tex]
Explanation:
From the question we are told that
The speed of the airplane is [tex]u = 92.3 \ m/s[/tex]
The angle is [tex]\theta = 51.1^o[/tex]
The altitude of the plane is [tex]d = 532 \ m[/tex]
Generally the y-component of the airplanes velocity is
[tex]u_y = v * sin (\theta )[/tex]
=> [tex]u_y = 92.3 * sin ( 51.1 )[/tex]
=> [tex]u_y = 71.83 \ m/s[/tex]
Generally the displacement traveled by the package in the vertical direction is
[tex]d = (u_y)t + \frac{1}{2}(-g)t^2[/tex]
=> [tex] -532 = 71.83 t + \frac{1}{2}(-9.8)t^2[/tex]
Here the negative sign for the distance show that the direction is along the negative y-axis
=> [tex]4.9t^2 - 71.83t - 532 = 0[/tex]
Solving this using quadratic formula we obtain that
[tex]t = 20.06 \ s[/tex]
Generally the x-component of the velocity is
[tex]u_x = u * cos (\theta)[/tex]
=> [tex]u_x = 92.3 * cos (51.1)[/tex]
=> [tex]u_x = 57.96 \ m/s[/tex]
Generally the distance travel in the horizontal direction is
[tex]D = u_x * t[/tex]
=> [tex]D = 57.96 * 20.06 [/tex]
=> [tex]D = 1162.7 \ m [/tex]
Generally the angle of the velocity vector relative to the ground is mathematically represented as
[tex]\beta = tan ^{-1}[\frac{v_y}{v_x } ][/tex]
Here [tex]v_y[/tex] is the final velocity of the package along the vertical axis and this is mathematically represented as
[tex]v_y = u_y - gt[/tex]
=> [tex]v_y = 71.83 - 9.8 * 20.06[/tex]
=> [tex]v_y = -130.05 \ m/s [/tex]
and v_x is the final velocity of the package which is equivalent to the initial velocity [tex]u_x[/tex]
So
[tex]\beta = tan ^{-1}[-130.05}{57.96 } ][/tex]
[tex]\beta =- 65.55^o[/tex]
The negative direction show that it is moving towards the south east direction