Answer:
, the minus meaning west.
Explanation:
We know that linear momentum must be conserved, so it will be the same before (
) and after (
) the explosion. We will take the east direction as positive.
Before the explosion we have 
.
After the explosion we have pieces 1 and 2, so 
.
These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.
Since we know momentum must be conserved we have:
Which means (since we want 
and 
):
So for our values we have:
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
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:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN
