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3 years ago

What is CHA-CHA-CHA.

Physics

1 answer:

polet [3.4K]3 years ago

5 0

Answer:

The cha-cha-cha, is a dance of Cuban origin. It is danced to the music of the same name introduced by Cuban composer and violinist Enrique Jorrin in the early 1950s. This rhythm was developed from the danzón-mambo

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143°C = _____ 416 K -130 K 0 K 143 K

dezoksy [38]

The answer is 0k because 143c equals nothing

5 0

2 years ago

Read 2 more answers

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

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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

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$