Answer:
2.The forces are unbalanced.
5.The net force is to the right.
6.The book is moving to the right.
Explanation:
correct on edge :)
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
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:
Part a)
Part b)
Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
now we will have force balance in Y direction
Part a)
so from above equations we have
Part b)
Now for tension in right string we will have
