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Liono4ka [1.6K]
3 years ago
13

When water is boiled at a pressure of 2.00 atm, the heat of vaporization is 2.20×106J/kg2.20×10

Physics
1 answer:
Grace [21]3 years ago
5 0

Answer:

Explanation:

a ) When 1 kg water is boiled at constant pressure of 1  atm , its volume increases by following volume

(.824 - .001 )m³

.823 m³

work done by steam  = increase in volume x pressure

.823 x 10⁵ J

Heat added

=  latent heat of vaporization x mass

= 2260000 J x 1

= 22.6 x 10⁵ J

Increase in internal energy of gas

= heat added - work done by gas

= (22.6 - .823) x 10⁵ J

= 21.777 x 10⁵ J .

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For 1983 through 1989, the per capita consumption
Tresset [83]

Answer:

y = 43.55 + 2.15t

Explanation:

We were told that in 1983, the per capita consumption was 37.1 pounds, and in 1989 it was 50 pounds.

If we assume t = 0 corresponds to year 1980. Then, for 1983 it will be t = 3 and for 1989,it will be t = 9.

Thus, expressing the information as ordered pairs, we have; (3,37. 1) and (9,50).

Let us now find slope of the linear function:

m1 = (y2 - y1)/(t2 - t1)

m1 = (50 - 37.1)/(9 - 3)

m1 = 2.15

So, we can find the linear equation as;

y - 37.1 = 2.15(t - 3)

y = 37.1 + 2.15t - 6.45

y = 43.55 + 2.15t

8 0
3 years ago
PLEASE HELP!
Anna [14]

Answer:

Explanation:

At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume

ΔV = 9.2 - 5.6 = 3.6 L

3.6 L = 3.6 x 10⁻³ m³

ΔV = 3.6 x 10⁻³ m³

P = 3.7 x 10³ Pa

So work done

= 3.7 x 10³ x 3.6 x 10⁻³ J

= 13.32 J .

( c ) is the answer , because work is done by the gas so it will be positive.

5 0
3 years ago
Read 2 more answers
What’s the opposite of an electron
Anna007 [38]
As its charge, proton -a positive charged molecule at the center of an atom- is the opposite of the electron -the particle which is orbiting the center of an atom.
7 0
3 years ago
3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.​
BigorU [14]

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a.  PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

6 0
3 years ago
Read 2 more answers
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the
mash [69]

Answer:

\Delta x = v_0 t + \frac{1}{2}at^2

Explanation:

To solve this problem, we can use the following suvat equation:

\Delta x = v_0 t + \frac{1}{2}at^2

where

\Delta x is the vertical displacement of the frog

v_0 is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from v_0, all the other quantities are known. In fact:

\Delta x =0.1 m is the vertical displacement

t = 2 s is the total time of flight

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

Therefore, solving for v_0, we find the initial velocity of the frog:

v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s

4 0
3 years ago
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