Compare and contrast aerobic physical activity with anaerobic physical activity.
2 answers:
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Answer:
2.36 x 10^6 J
Explanation:
Tc = 0°C = 273 K
TH = 22.5°C = 295.5 K
Qc = heat used to melt the ice
mass of ice, m = 85.7 Kg
Latent heat of fusion, L = 3.34 x 10^5 J/kg
Let Energy supplied is E which is equal to the work done
Qc = m x L = 85.7 x 3.34 x 10^5 = 286.24 x 10^5 J
Use the Carnot's equation
QH = 309.8 x 10^5 J
W = QH - Qc
W = (309.8 - 286.24) x 10^5
W = 23.56 x 10^5 J
W = 2.36 x 10^6 J
Thus, the energy supplied is 2.36 x 10^6 J.
Answer:
(a) 19.71801m/s Velocity just before going up the ramp.
(b) 74.56338m.
Explanation:
We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.
Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.
Following formulas will be used.
(a) Calculating velocity right before going up the ramp.
Wooden block is going on a straightaway and has net for on it.
=
and this force produces acceleration of
With this acceleration, wooden block would reach at the foot of ramp in.
and final velocity will be
this velocity of wooden box just before going up the ramp.
(b) How far up the ramp will the wooden block go before stopping.
Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.
and deceleration along the ramp is
Where force of friction along the inclined plane.
is a component of g along normal of the inclined plane.
And with this deceleration time needed to get wooded block to stop is.
and in that time wooden block would travel
This is how up wooden box will go before coming to stop.