Which kind of air front forms when a warm air mass meets the area of a cooler air mass?

Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?

nikitadnepr [17]

Answer:

The appropriate response is "Optical printer ".

Explanation:

A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses.
A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.

4 0

3 years ago

Mixed powders may be categorized as

Arada [10]

Answer:A powder is an assembly of dry particles dispersed in air. If two different powders are mixed perfectly, theoretically, three types of powder.

Explanation:

4 0

2 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = 315.876 m/s

the root mean square velocity of the oxygen gas is

478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank

6 0

2 years ago

Wind tends to move from pressure areas. low, high high, high low, low high, low

andrew-mc [135]

Answer:

high, low

Explanation:

Energy always flows from a higher level to a lower level.
It is analogous to the waterfall where waterfalls from a higher level to a lower level.
So in the case of the pressure of the gas, when there are any numbers of molecules in a given volume of space. The gas is said to be at high pressure.
When there are fewer molecules in the given volume. The gas is said to be at lower pressure.
Due to a large number of atoms, the high-pressure gas exerts more force on the container than the force exerted by the low-pressure gas.
If a hose is connected between these two containers, gas rushes from high pressure to the low pressure. Since the force exerted by the high-pressure gas is greater than that of low-pressure gas.

So, the wind tends to move from high-pressure areas to low pressure.

3 0

2 years ago

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

2 years ago