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forsale [732]
3 years ago
8

Summarize the weather conditions related to a warm front. Remember to include all data collected on warm fronts in this activity

to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc.).
Physics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

to a warm front. Remember to include all data collected on warm fron … ... Remember to include all data collected on warm fronts in this activity to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc. 1

Explanation:

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

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

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\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

\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

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

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

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
.<br> Why are meteorites moving?
lisov135 [29]

Answer:

Glow

Explanation:

Actually, it is the air in front of the meteoroid that heats up. The particle is traveling at speeds between 20 and 30 kilometers per second. It compresses the air in front, causing the air to get hot. The air is so hot it begins to glow — creating a meteor - the streak of light observed from Earth.

Hope this helped!

7 0
3 years ago
Can someone help me ASAP
Phantasy [73]

Well I don't know.  Let's actually LOOK at the picture and see if that helps.

A,  B,  C,  and D all have the same TOTAL length, but  A  has the most waves crammed into that same total length.

By golly, that means the length of <u><em>each</em></u> wave in  A  must be shorter than each wave in  B,  C,  or D.

The correct choice is <em> A </em>.  Looking at the picture did the trick !

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3 years ago
What building materials do you believe would work well to build a home in that area? Explain why you chose these materials.
Luden [163]
Depends on what the area is. If it’s a rural place, Wood is cheep & easy to build. If there’s a lot of corrosion, strong weather/hurricane, bricks.
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4 years ago
A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn
olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0
4 years ago
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