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
To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency
At the same time we have that
Therefore there is not have same units and both are not consistent and the correct answer is B.
Instantaneous velocity is the velocity at a specific instant in time. I bet you are taking Honors Physics.
Answer: 0.0146m
Explanation: The formula that defines the velocity of a simple harmonic motion is given as
v = ω√A² - x²
Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.
The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA
One half of maximum speed = speed of motion
3ωA/2 = ω√A² - x²
ω cancels out on both sides of the equation, hence we have that
A/2 = √A² - x²
(0.0169)/2 = √(0.0169² - x²)
0.00845 = √(0.0169² - x²)
By squaring both sides, we have that
0.00845² = 0.0169² - x²
x² = 0.0169² - 0.00845²
x² = 0.0002142
x = √0.0002142
x = 0.0146m
Answer:
Acceleration
Explanation:
Its speed or velocity change
