Ask question

Ask question

All categories

oksano4ka [1.4K]

3 years ago

9

If it required 575 cal to convert a sample of liquid water to a vapor, how many calories would be released when the vapor conden

sed back to a liquid?
A.
more than 575 cal
B.
less than 575 cal
C.
575 cal

1 answer:

Maru [420]3 years ago

5 0

B because when a liquid converts to a vapor is it going to be less

You might be interested in

Density is 1.3kg/m3 volume=? mass=? 80m 40m 15m<br>

mash [69]

From the given information, it's not possible to find the mass or the volume of the sample.

Density is their RATIO, but you can't tell if the sample is the size of an ant or a school-bus.

4 0

2 years ago

The average speed of a nitrogen molecule in air is about 6.70 ✕ 102 m/s, and its mass is about 4.68 ✕ 10-26 kg.(a) If it takes 3

Sveta_85 [38]

Answer:

a = 4.32*10^15 m/s^2

Explanation:

In order to calculate the acceleration of the nitrogen molecule, you take into account that the force experienced by the molecule, is equal to the change in its momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the nitrogen molecule

Δv: change in the speed of the molecule

Δt: time lapse of the change of the momentum of the molecule = 3.10*10^-13 s

Furthermore, by the Newton second law, you have:

$F=ma=m\frac{\Delta v}{\Delta t}\\\\a=\frac{\Delta v}{\Delta t}$ (2)

The change in the speed is given by:

$\Delta v=v-v_o=6.70*10^2m/s-(-6.70*10^2m/s)=1.34*10^3\frac{m}{s}$

where you have taken into account the direction of the initial and final speed.

You replace the values of all parameters in the equation (1) in order to calculate the acceleration:

$a=\frac{1.34*10^3m/s}{3.10*10^{-13}s}=4.32*10^{15}\frac{m}{s^2}$

The acceleration of the nitrogen molecule is 4.32*10^15m/s^2

5 0

3 years ago

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

Dennis_Churaev [7]

Answer:

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is $m_b = 0.490kg$

The distance of compression is $x$

The force constant is $k = 450 N/m$

The radius of the circular track is $R =1.00m$

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

$0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

$W_f = 7 * 3.142$

$= 21.99 J$

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

$PE = m_bgh$

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

$PE = 0.490 * 9.8 *2$

$=9.604 J$

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

8 0

3 years ago

Can someone help me please

enyata [817]

Answer:

I think its distance

Explanation:

when measuring how far a p.o art u can use mm

3 0

3 years ago

Read 2 more answers

Why smartphone and computer are called modern technology?

topjm [15]

Answer:

Hey

I am sure from what i know that your answer should be the following:

Modern technology is deeply tied to efficiancy and speed. Think of the knew 5G network it gives us very fast internet, or perhaps our very nessisitys such as our phone or cumputer. On a phone you have so much (thats to say you can acces alot easly); pictures videos dictionarys video games etc etc.

6 0

3 years ago