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Art [367]
3 years ago
7

What are some predicted affects of climate change linked to global warming

Physics
1 answer:
Alla [95]3 years ago
5 0
Decrease in atmospheric ozone, which results in lower protection from various stellar particles
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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
I will mark you brainlist!
kirill115 [55]
Tornado- Trees knocked down, debris everywhere, ground and dirt scattered.
7 0
3 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
kati45 [8]

Answer:

The  coefficient of kinetic friction  \mu_k =  0.724

Explanation:

From the question we are told that

   The  length of the lane is  l =  36.0 \  m

    The speed of the truck is  v  =  22.6\  m/s

     

Generally from the work-energy theorem we have that  

    \Delta KE  =   N  *  \mu_k *  l

Here N  is the normal force acting on the truck which is mathematically represented as

     \Delta KE is the change in kinetic energy which is mathematically represented as

        \Delta KE =  \frac{1}{2} *  m *  v^2

=>     \Delta KE =  0.5  *  m *  22.6^2

=>      \Delta KE =  255.38m

        255.38m =    m *  9.8  *  \mu_k *   36.0

=>     255.38  =    352.8  *  \mu_k

=>   \mu_k =  0.724

 

6 0
3 years ago
3 3
olganol [36]

Answer:

I Dont know u answer it

Explanation:

8 0
2 years ago
The initial velocity of a micro van is 15 m/s. It gains a velocity of 40 ms in 10 seconds. Calculate the average velocity and ac
PSYCHO15rus [73]

Answer:

{ \bf{average \: velocity =  \frac{15 + 40}{2}}} \\   = 27.5 \:  {ms}^{ - 1}  \\  { \bf{acceleration =  \frac{v - u}{t} }} \\  =  \frac{40 - 15}{10}  \\  = 2.5 \:  {ms}^{ - 2}  \\  \\ { \tt{second \: qn : }} \\ { \bf{final \: velocity =u + at }} \\ v = 0 + (5 \times 10) \\  = 50 \:  {ms}^{ - 1}

5 0
2 years ago
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