All categories

3 years ago

In what changes of state do atoms lose energy? Check all that apply

1 answer:

5 0

In this case, Sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.

You might be interested in

In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretio

Studentka2010 [4]

As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.

Now the process:

First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.

After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.

As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.

4 0

3 years ago

17.

Zolol [24]

Answer:

I don't know this answer at all

Explanation:

I don't know about these problems

6 0

3 years ago

A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling

Tasya [4]

Answer:

Solving for time :

(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known

2. The unknown constant that we want to solve)

s = (1/2)(u+v)t <--- one of the formulas

from linear motion

s (distance) = 0.05m

u (initial velocity) = 100m/s

v (final velocity) = 0 m/s (it stops)

t (time taken for change in velocity) = to be found

0.05 = (1/2)(100+0)t

t = 0.001 seconds

Solving for the resistant force :

Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.

When the bullet stops :

F net = 0

F r = F imp

F r = (mu -mv)/t

F r = (0.01x100-0.01x0)/0.001

F r = 1/0.001

F r = 1000N

4 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$