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3 years ago

What makes a funnel appear black

Physics

1 answer:

fredd [130]3 years ago

6 0

Answer:

Air

Explanation:

The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.

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To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago

A striped billiard ball moves toward the right with speed 3 m/s. A solid billiard ball with the same mass moves toward the left

Helen [10]

Answer:

Final speed of striped ball is 3 m/s in left direction .

Explanation:

Given :

Two billiard ball with the same mass moves toward the left at the same speed 3 m/s .

Let , us assume right hand side direction to be positive and left hand side direction to be negative .

Also , let speed of ball after collision is (striped ball ) u and (solid ball) v .

It is also given that the collision is elastic .

Therefore , kinetic energy is conserved .

$\dfrac{m(3)^2}{2}+\dfrac{m(3)^2}{2}=\dfrac{mu^2}{2}+\dfrac{mv^2}{2}\\\\u^2+v^2=18$ ...... ( 1 )

Also , by conserving linear momentum .

We get :

$3m-3m=mu+mv\\u=-v$ ...... ( 2 )

Putting value of u from equation 2 to equation 1 .

We get :

$2v^2=18\\v=3\ m/s$

And , u = -3 m/s .

Therefore , final speed of striped ball is 3 m/s in left direction .

Hence , this is the required solution .

4 0

3 years ago

If 5000Pa a pressure is exerted on an object with the contact area 0.04m^2. Calculate the mass of an object

Whitepunk [10]

Pressure = 5000 Pa
Contact Area = 0.04 m^2
Acceleration due to gravity = 9.8 m/s^2
Let the force be F.
We know, Force = Pressure × Contact Area
Therefore, Force = 5000 Pa × 0.04 m/s^2
or, Force = 200 N
We know, force = mass × acceleration
Therefore, mass = force ÷ acceleration
or, mass = 200 N ÷ 9.8 m/s^2 = 20.4 Kg

Answer:

20.4 Kg

Hope you could understand.

If you have any query, feel free to ask.

6 0

2 years ago

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the h

Nataly_w [17]

Answer:

v₁ = 1,606 10⁴ m / s

Explanation:

For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂

The pressure when the water comes out is the atmospheric pressure

P₁ = P_atm

The difference in height between the street and the nozzle on the stairs is

y₂-y₁ = 15 m

Now let's use the continuity equation

v₁ A₁ = v₂ A₁

The area of a circle is

A = π r² = π (d/2)²

v₁ π d₁²/ 4 = v₂ π d₂²/ 4

v₂ = v₁ d₁² / d₂²

Let's replace

P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0

P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]

v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

Let's reduce the magnitudes to the SI system

d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

d₁ = 2.02 in = 5.13 10⁻² m

Let's calculate

v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2

v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

v₁ = √ 2.2477 10⁸ /0.8710

v₁ = 1,606 10⁴ m / s

4 0

2 years ago

Discuss one way in which global warming aggravates the effect of radiation

alukav5142 [94]

The energy that then radiates out from the surface, longwave radiation, is trapped by the same greenhouse gases, warming the air, oceans, and land. This process, appropriately dubbed “the greenhouse effect,” is how global warming occurs.

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2 years ago