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Zinaida [17]
3 years ago
8

Assuming standard pressure, what physical constants are needed to calculate the change in enthalpy for raising a sample of water

from 250 K to 300 K. I) the heat capacity of ice II) the enthalpy of fusion of water III) the heat capacity of water IV) the enthalpy of vaporization of water V) the heat capacity of steam
Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

I) Heat capacity of ice (Sensible, ice), II) Enthalpy of fusion of water (Latent, fusion) and III) Heat capacity of water (Sensible, water).

Explanation:

At standard pressure, water has a fusion point and boiling point of 273.15 K and 373.15 K, respectively. Then, the total change in enthalpy is the sum of two sensible indicators and one latent indicator. That is:

\Delta H_{total} = \Delta H_{sensible, ice} + \Delta H_{latent, fusion} + \Delta H_{sensible, water}

The needed physical constants are: Heat capacity of ice (Sensible, ice), Enthalpy of fusion of water (Latent, fusion) and Heat capacity of water (Sensible, water).

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A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s
Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

  • Time of flight = t = 20 s
  • Angle of the initial velocity of projectile with the horizontal = \theta = 30^\circ

<u>Assume:</u>

  • Initial velocity of the projectile = u
  • R = Range of the projectile during the time of flight
  • H = maximum height of the projectile
  • D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

  • Initial horizontal velocity of the projectile = u\cos \theta
  • Initial horizontal velocity of the projectile = u\sin \theta

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0
2 years ago
Explain what parallel medium boundaries do to the path of the wave after it travels through both boundaries. (air and water)
Brilliant_brown [7]

Answer:

The boundaries cause the waves to change direction an effect called <u>refraction.</u>

Explanation:

When a wave crosses a boundary between different materials, the speed of the wave and its wavelength changes.When passing from air to water the two properties (speed and wavelength) decreases, and the wave is observed to change direction as it crosses the boundary between the two material.The bending of the wave is called refraction.

7 0
3 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
2 years ago
A group of scientists have obtained some experimental results.
topjm [15]
<span>In order best to find out whether the obtained experimental results are worth mor etime and resources the group of scientists should present their results (could be done also in poster session) to other scientists in order to hear their opinion and get a feedback. The shoudl also ask another researchers to redo the experiments and to compare the results. </span>
4 0
3 years ago
Read 2 more answers
Which of the following ranching practices contributes to soil erosion?
Natasha2012 [34]

A. overgrazing of livestock

Explanation:

Overgrazing of livestock is the ranch practice that contributes to soil erosion.

Soil erosion is the washing away of the top layer of the earth on which plant grows.

  • Plants are covers of the earth and they prevent agents of denudation from actively washing away the soil.
  • When livestock are introduced into a place, continuous grazing leaves the surface bare.
  • The plants becomes depleted and the surface is exposed.
  • This fast tracks the process of overgrazing.

learn more:

Soil erosion brainly.com/question/2473244

#learnwithBrainly

6 0
3 years ago
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