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3 years ago

Giving 25 points need help ASAP!

Physics

1 answer:

Bumek [7]3 years ago

4 0

Answer:

i dont exactly know the answer but i do know the formula

Explanation:

1/ 1/R1 + 1/ R2+

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What is the meaning of physic

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the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

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3 years ago

Compare these two waves.

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3 years ago

Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the

murzikaleks [220]

Answer:

The function has a maximum in $x=3$

The maximum is:

$f(3) = 39$

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

$f(x)' = -4*2x + 24=0$

$-4*2x + 24=0$

$8x=24$

$x=3$

Now find the second derivative of the function and evaluate at x = 3.

If $f (3) ''< 0$ the function has a maximum

If $f (3) '' >0$ the function has a minimum

$f(x)''= 8$

Note that:

$f(3)''= -8$

the function has a maximum in $x=3$

The maximum is:

$f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39$

4 0

3 years ago

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

HIII DROPPING COINS YALL

suter [353]

Answer:

tysm

Explanation:

8 0

3 years ago

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