Where do seasons alternate between a very dry season and a monsoon season, with a stable warm climate?

What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

ratelena [41]

= 80 N/40 kg
= 2 m/s 2

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

A protostar is _____.

defon

<u>Answer</u>

5.the stage before a star becomes a main sequence star

<u>Explanation</u>

A protostar is a small star that is still gathering its masses. When it forms enough masses it make a parent molecular cloud.

This been the case, from the choices given, the correct statement about a protostar is;

5.the stage before a star becomes a main sequence star

8 0

3 years ago

Read 2 more answers

How can citizen -scientists help with the prediction of future earthquakes

iVinArrow [24]

Answer:

Avoid downed power lines and stay away from buildings and bridges from which heavy objects might fall during an aftershock. Stay away until local officials tell you it is safe. A tsunami is a series of waves that may continue for hours. Do not assume that after one wave the danger is over.

3 0

3 years ago

When converted to a household measurement, 9 kilograms is approximately equal to a

kolezko [41]

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

7 0

3 years ago