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
A type of matter made up of a combination of elements is called a compound. for example table salt, is made up of two elements that by themselves are lethal but chemically combined cancel each other out there for becoming edible.
Answer:
a) 5 N b) 225 N c) 5 N
Explanation:
a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:
F₁₂ = K Q₁ Q₂ / r₁₂²
So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N
b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.
So, we will have F₁₂ = 9. 25 N = 225 N
c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:
F₁₂ = 25 N .1/5 = 5 N
Answer:
I would say A but am not sure
