Which of the following processes forms clouds A. Boiling B.Condensation C.Freezing D.Precipitation

An apple is whirled round in a horizontal circle on the end of a string which is tied to the stalk. It is whirled faster and fas

jek_recluse [69]

when the apple moves in a horizontal circle, the tension force in the string provides the necessary centripetal force to move in circle. the tension in the string is given as

T=mv²/r

where T = tension force in the string , m = mass of the apple

v = speed of apple , r = radius of circle.

clearly , tension force depends on the square of the speed. hence greater the speed, greater will be the tension force.

at some point , the speed becomes large enough that it makes the tension force in the string becomes greater than the tensile strength of the string. at that point , the string breaks

6 0

3 years ago

The unusually bright centers found in some galaxies are called ______.

Nimfa-mama [501]

<span>Active Galactic Nuclei.</span>

4 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

The average human has a density of 945 kg/m3 after inhalation, and 1020 kg/m3 after exhalation. Fresh water has a density of 100

Montano1993 [528]

Answer:

a. The human body has nearly the same density as salt water after exhaling.

b. The human body will always float in the Dead Sea.

Explanation:

According to the concept of floating on the basis of density, any body that is put in a fluid of density greater than its own density will always float due to the force of buoyancy from the liquid.

The portion of the object submerged while the object is floating depends upon the density of the object as compared to the density of the fluid. This is governed by the equation:

$\rho_f.V_s=\rho_o.V_o$

where:

$\rho_f=$ density of the fluid

$\rho_o=$ density of the object

$V_s=$ volume of the object submerged in the fluid

$V_o=$ total volume of the object

6 0

4 years ago

I was an American inventor and businessman who is best known for inventing the electric light bulb. Who am I?

wel

You are Thomas Alva Edison. You also invented the phonograph
and the first practical movie camera. Sadly, you died almost exactly
9 years before I was born.

4 0

3 years ago

Read 2 more answers