How much work can be done by a 20 W motor in 5 seconds?

Anton [14]

A light-year is how astronomers measure distance in space.
-Light can travel 300,000 kilometres in one seconds!
Light year is the distance that light travel can travel in one year.
-Light can also travel 9,00,000,000,000 kilometers.
-One light is about 10 trillion km.
-Almost 300 years ago ,a Danish scientist named olaus roemer,was the fist man to estimate the distance light travels in a year -and using that he figure out the approximate speed of light!

4 0

3 years ago

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Calculate their densties in Si unit. 200mg,0.0004m

Kryger [21]

Question: calculate their densties in Si unit.

200mg,0.0004m³

Answer:

0.5 kg/m³

Explanation:

Applying,

D = m/V........................ Equation 1

Where D = density, m = mass, V = volume.

From the question,

Given: m = 200 mg = (200/1000000) kg = 2.0×10⁻⁴ kg, V = 0.0004 m³ = 4.0×10⁻⁴ m³

Substitute these values into equation 1

D = (2.0×10⁻⁴ kg)/(4.0×10⁻⁴)

D = 2/4

D = 0.5 kg/m³

Hence the density in S.I unit is 0.5 kg/m³

3 0

2 years ago

Sand dunes during the Dust Bowl are an example of:

choli [55]

Sand dunes during the Dust Bowl are an example of windward.

The sand dunes are formed due to high movement of wind in the deserts which make dunes in the direction where wind is moving. Windward refers to facing the wind or situated on the side facing the wind so in those regions where the wind moves bring sand with itself and make dunes in that area so we can conclude that sand dunes are formed in the windward regions.

Learn more: brainly.com/question/24879138

5 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

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