Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
The initial temperature is 
The mass of the tanker with petrol is 250000 N.
We are given that,
Mass of tanker= m =2800 kg
Volume of petrol= v =30 m³
Density of petrol= d =740 kg/m³
Thus , mass , density and volume relation can be given as,
density= mass/ Volume
Mass = Density × Volume
Mass = 740× 30
Mass = 22200 kg
The mass of the petrol is 22200 kg.
Total mass of tanker with petrol = Mass of petrol + Mass of tanker
Total mass of tanker with petrol= 22200+ 2800 kg
Total mass of tanker with petrol= 25000 kg
Total weight of the tanker with petrol = Mass of tanker full of petrol× g
Where, weight = m × g ,(g =10m/s²)
Total weight of the tanker with petrol= 25000×10 = 250000 N
Therefore, the mass of petrol, total mass of tanker with petrol and weight of tanker with petrol would be 22200 kg, 25000 kg and 250000 N.
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That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
Answer:
0.5 m/s north
Explanation:
Take east to be +x, west to be -x, north to be +y, and south to be -y.
His displacement in the x direction is:
x = 20 m − 20 m = 0 m
His displacement in the y direction is:
y = 10 m
His total displacement is therefore 10 m north.
His velocity is equal to displacement divided by time.
v = 10 m north / 20 s
v = 0.5 m/s north
