All categories

2 years ago

Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain

1 answer:

8 0

The surrounding air will become warm when water vapor condenses. The vapors when become water will give away latent heat they have, we know that latent heat is required for the object to change states, so, the latent heat the water vapor had when it became water vapor from water will be given out when it again becomes water.

You might be interested in

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

2 years ago

A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of

Crank

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate 0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant

the kinetic energy for 60 mph larger than this of 30 mph

= $\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}$

= $\dfrac{v_1^2}{v_2^2}$

= $\dfrac{60^2}{30^2}$

= 4

gain in kinetic energy = P x t

time = 4 x 1.5

= 6 s

8 0

2 years ago

Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r

Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling $P_{xi}$ the initial momentum of object X and $P_{yi}$ the initial momentum of object Y, we can derive the total initial momentum of the system: $P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}$

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: $M * v_f=400kg * v_f$