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2 years ago

Science If you want to find the energy quantum of light, you multiply the frequency of the radiation (v) by "h". What is "h"?

Physics

1 answer:

weqwewe [10]2 years ago

7 0

Answer:

"h" signifies Planck's constant

Explanation:

In the equation energy E = h X v

The "h" there signifies Planck's constant

Planck's constant is a value, that shows the rate at which the energy of a photon increases/decreases, as the frequency of its electromagnetic wave changes.

It was named after Max Planck who discovered this unique relationship between the energy of a light wave and its frequency.

Planck's constant, "h" is usually expressed in Joules second

Planck's constant = $6.62607015 \times 10^{-34} J.s$

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A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

2 years ago

Which of the following is a definition of glass transition temperature?

aleksandr82 [10.1K]

Answer:

c. The temperature at which a glass transforms from a solid to liquid.

Explanation:

The glass transition temperature is said to be a temperature range when a polymer structure transition from a glass or hardy(solid) material to a rubber like or viscous liquid material.

The glass transition temperature is an important property that is critical in product design.

5 0

3 years ago

A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl

KengaRu [80]

Answer:

$a) \ d_2=d_1\\b) \ d_2=4d_1$

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

$E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}$

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0

3 years ago

How does the sun rise and set

MatroZZZ [7]

Answer:

c. earth rotates on it tilted axis and orbit around the sun at the same time

Explanation:

Earth rotation can be defined as the amount of time taken by planet earth to complete its spinning movement on its axis.

This ultimately implies that, the rotation of earth refers to the time taken by earth to rotate once on its axis. One spinning movement of the earth on its axis takes approximately 24 hours to complete with respect to the sun. Thus, this makes us to experience day and night (sun rise and sun set) due to the rotation of planet about its axis.

Hence, the sun rise and set because the earth rotates on it tilted axis and orbit around the sun at the same time

8 0

3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

6 0

3 years ago