Study of interaction between matter and energy using mechianical description
Winter and Summer happen every year, pretty much on schedule.
When they happen, the 'weather' changes, but the 'climate' doesn't.
If you live in the right place, then you might get several blizzards every
Winter. They're changes in the 'weather' but not in the 'climate'.
If the temperature where you live stays well below zero for a hundred
years, then you might be entering an ice age.
Or let's say your town has had blizzards every Winter for the past 20 years
but it never had one for 100 years before that.
Either of these would be a climatic change.
Answer:
1.2ms⁻²
Explanation:
F = ma
12 = 10*a
12/10 = a
6/5 ms⁻² = a or 1.2 ms⁻² = a
Answer:
c = 4
Explanation:
From work-energy theorem KE = workdone.
Given F = (cx - 3.00x²)i
W = ∫Fdx = ∫(cx - 3.00x²)dx = cx²/2 –3.00x³/3 + A
W = cx²/2 –x³ + A
Where A is a constant
At x = 0, KE = 20J
So W = 20J at x = 0
20 = c×0 - 0 +A
A = 20
So W = cx²/2 –x³ + 20
Also when x = 3.00m, W = KE = 11J
So
11 = c×3²/2 – 3³ + 20
11 = 4.5c – 7
4.5c = 11 + 7
4.5c = 18
c = 18/4.5 = 4
c = 4
<h3>
Answer:</h3>
539.56 Joules
<h3>
Explanation:</h3>
- Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
- Efficiency = (work output/work input) × 100%
- Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.
In this case;
Efficiency = 94%
Work input = 574 Joules
Therefore, Assuming work output is x
94% = (x/574 J) × 100%
0.94 = (x/574 J)
<h3>x = 539.56 J</h3>
Thus, you get work of 539.56 J from the machine
