Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
4. Grass - Caterpillar - Hedgehog - Fox
5. Caterpillar, Rabbit, Mouse.
6. Cougar and Fox.
7. Bacteria
8. The bird, hedgehog, Fox and cougar would be effected since the Hedgehogs and birds would soon die out due to the loss of their food. Once they die out, the cougar and Fox would have no predators left to eat.
Answer:
W = - 118.24 J (negative sign shows that work is done on piston)
Explanation:
First, we find the change in internal energy of the diatomic gas by using the following formula:
where,
ΔU = Change in internal energy of gas = ?
n = no. of moles of gas = 0.0884 mole
Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)
Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K
ΔT = Rise in Temperature = 18.8 K
Therefore,
Now, we can apply First Law of Thermodynamics as follows:
where,
ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)
W = Work done = ?
Therefore,
<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>
I need pictures or something
