Answers
Answer:
The meteoroid's speed is 18.5 km/s
Explanation:
Given that,
Speed = 14.8 km/s
Distance
We need to calculate the meteoroid's speed
The total initial energy
Where, m = mass of meteoroid
G = gravitational constant
=mass of earth
r = distance from earth center
Now, The meteoroid hits the earth then the distance of meteoroid from the earth 's center will be equal to the radius of earth
The total final energy
Where,
=radius of earth
Using conservation of energy
Put the value of initial and final energy
Put the value in the equation
Hence, The meteoroid's speed is 18.5 km/s
Final answer:
To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. The final velocity of the meteoroid is approximately 13.4 km/s.
Explanation:
To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. Since there is no air friction, the mechanical energy of the meteoroid is conserved as it falls towards Earth. The initial kinetic energy of the meteoroid is equal to the final kinetic energy plus the gravitational potential energy.
First, we find the initial kinetic energy of the meteoroid using the formula KE = (1/2)mv^2, where m is the mass of the meteoroid and v is its initial velocity relative to the center of the Earth. Since the mass is not given, we can assume it cancels out in the equation.
Next, we calculate the gravitational potential energy of the meteoroid using the formula PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height from which the meteoroid fell. The height can be calculated by subtracting the radius of the Earth from the distance from the center of the Earth to the moon's orbit (h = 3.84 × 10^8 m - 6.37 x 10^6 m).
Solving for the final velocity, we equate the initial kinetic energy and the sum of the final kinetic energy and gravitational potential energy. Rearranging the equation, we find that the final velocity is the square root of (initial velocity squared minus 2 times g times h).
Plugging in the given values, the final velocity of the meteoroid as it hits the Earth is approximately 13.4 km/s.
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Related Questions
A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?
A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.
What is it called when a Rock forms due to heat and pressure in the earth?
A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?
B. At the tip of a blade, what is the centripetal acceleration?
C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?
Answers
A) The speed of the tip of the blade is 76.2 m/s
B) The centripetal acceleration of the tip of the blade is
D) The force required to keep the droplet moving in circular motion is 0.39 N
E) The ratio of the force to the weight of the droplet is 4.0
Explanation:
A)
We know that the blade of the turbine is rotating at an angular speed of
First, we have to convert this angular speed into radians per second. Keeping in mind that
We get
The linear speed of a point on the blade is given by:
where
is the angular speed
r is the distance of the point from the axis of rotation
For a point at the tip of the blade,
r = 56 m
Therefore, its speed is
B)
The centripetal acceleration of a point in uniform circular motion is given by
where
v is the linear speed
r is the distance of the point from the axis of rotation
In this problem, for the tip of the blade we have:
v = 76. 2 m/s is the speed
r = 56 m is the distance from the axis of rotation
Substituting, we find the centripetal acceleration:
D)
The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:
where
m is the mass of the droplet
v is the linear speed
r is the distance from the centre of rotation
The data in this problem are
m = 10 mg = 0.010 kg is the mass of the droplet
v = 2.5 m/s is the linear speed
r = 16 cm = 0.16 m is the radius of the circular path
Substituting,
E)
The weight of the droplet is given by
where
m = 10 mg = 0.010 kg is the mass of the droplet
is the acceleration of gravity
Substituting,
The force that keeps the droplet in circular motion instead is
F = 0.39 N
Therefore, the ratio between the two forces is
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Answers
Answer:
E=930.84 N/C
Explanation:
Given that
I = 1150 W/m²
μ = 4Π x 10⁻⁷
C = 2.999 x 10⁸ m/s
E= C B
C=speed of light
B=Magnetic filed ,E=Electric filed
Power P = I A
A=Area=4πr² ,I=Intensity
E=930.84 N/C
Therefore answer is 930.84 N/C
Final answer:
To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.
Explanation:
To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.
Subbing in the given values, we can compute Em as:
Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]
This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.
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where Bx = 3.3 X 10-6 T, By = 3.9 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.
1)
What is f, the frequency of this wave?
GHz
2)
What is I, the intensity of this wave?
W/m2
3)
What is Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0?
W/m2
4)
What is Ex, the x-component of the electric field at (x = 0, y = 0, z = 0) at t = 0?
V/m
5)
Compare the sign and magnitude of Sz, the z-component of the Poynting vector at (x=y=z=t=0) of the wave described above to the sign and magnitude of SIIz, the z-component of the Poynting vector at (x=y=z=t=0) of another plane monochromatic electromagnetic wave propagating through vaccum described by:
B? =(BIIxi^?BIIyj^)cos(kz??t)
where BIIx = 3.9 X 10-6 T, BIIy = 3.3 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.
SIIz < 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)
SIIz < 0 and magnitude(SIIz) = magnitude(Sz)
SIIz > 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)
SIIz > 0 and magnitude(SIIz) = magnitude(
Answers
Final answer:
The question involves computation of frequency, intensity, Poynting vector and electric field of an electromagnetic wave, and comparison between two such waves. The solutions result in approximately: 10 GHz for frequency, 3.07 x 10^-12 W/m^2 for intensity, 1.3 X 10^-19 W/m^2 for the z-component of Poynting vector, and 1.43 V/m for the electric field. Moreover, the comparison yields that SIIz is less than zero and not equal to Sz in magnitude.
Explanation:
The subject of your question relates to
electromagnetic waves
and their properties such as frequency, intensity, Poynting vector, and the electric field component. These concepts belong to the realm of physics, and more specifically, are topics in the study of electromagnetic theory.
To solve your questions:
- The frequency f can be found using the formula: f = c/λ where c is the speed of light in vacuum (~3x10^8 m/s). For λ = 3 cm or 0.03 m, computation yields f ≈ 10^10 Hz or 10 GHz.
- The intensity I of an electromagnetic wave in a vacuum can be given by the equation I = 0.5*c*ε0*E^2, where E is the electric field amplitude. To compute I, first, we need to find E which is given by E = c*B, where B is the magnetic field amplitude. Here, B is the square root of (Bx^2 + By^2) resulting in approximately 4.77*10^-6 T. Thus, E ≈ 1.43 V/m and solving for I gives us I ≈ 3.07 x 10^-12 W/m^2.
- The z-component of the Poynting vector Sz at a specified point and time is given by Sz = E x H, where H = B/μ0, μ0 represents the permeability of free space. At t = 0, Sz = Ex*Hy - Ey*Hx = Ex*By, resulting in Sz ≈ 1.3 x 10^-19 W/m^2.
- The x-component of the electric field at t = 0 Ex≈1.43 V/m.
- Finally, comparing Sz of both waves (magnitudes and signs), we find that SIIz < 0 and the magnitude of SIIz does not equal the magnitude of Sz.
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Final answer:
The frequency of the wave is 10 GHz. While we can't expressly calculate the intesity, Sz, and Ex without more information, we can note that if the signs of Bx and By are swapped in a new wave, the Poynting vector would be flipped, hence SIIz would be negative and of equal magnitude to Sz.
Explanation:
An electromagnetic wave propagating through vacuum is described by certain electromagnetic fields which are associated with frequency, intensity, and Poynting vector which indicates the direction of energy flow. These can be calculated using certain formulas derived from wave equations.
Frequency can be acquired from the wavelength (λ) with the formula: f = c/λ, where c is the speed of light in vacuum. Using given λ = 3 cm, we get f = 10^10 Hz or 10 GHz.
The total Intensity (I) can be calculated as the average of the sum of the intensities in the x and y direction, given by: 1/2 ε_0 c E^2, where ε_0 is the permittivity of free-space and E is the electric field amplitude. However, more information might be needed to calculate this value. Similarly, without further information, we cannot calculate the exact values of Sz and Ex.
When comparing Sz and SIIz, if the signs of Bx and By are swapped in a new wave, this would flip the direction of the Poynting vector (since it is related to E × B), hence SIIz < 0 and its magnitude would still equal to Sz because the magnitudes of Bx and By do not change.
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helpful to represent this motion?
A. stacking blocks to build a tower
B. freezing water in an ice cube tray
C. bouncing elastic balls off of each other and
the walls of a room
D. placing a closed, water-filled plastic bag in
the sun and watching condensation form
Answers
Answers
Answer:
0.5 lambda(wavelength)
Explanation:
We know that
The first harmonic for both side open ended pipe is
L= 1/2lambda
So L = 0.5*wavelength
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Answers
Answer:
a)Δy = 81.7mm
b)Δy = 32.7cm
Explanation:
To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:
where D is the separation between the slits and the screen where the interference pattern is observed.
a) In this case:
Δy = |y1max (λ1) − y1max (λ2)|
Δy =
Δy =
Δy =
Δy =
Δy = 81.7mm
The separation between these maxima is 81.7 mm
b)
Δy = |y₂max (λ1) − y₂max (λ2)|
Δy =
Δy =
Δy = 32.7cm
The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.
Final answer:
We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.
Explanation:
The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.
To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.
So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.
And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.
Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.
Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.
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