They require a medium to travel through
Answer:
104 N
Explanation:
m = 1300 kg
a = 0.08m/s^2
F = 1300*0.08
F = 104 N
Newtons is the unit of force.
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
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

The wavelength of the 2nd harmonic is:

The wavelength of the 4th harmonic is:

It is not possible to find any integer n such that
, therefore the correct options are A, B and D.
Answer:
Force, 
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

Finally, it stops, v = 0



F = -22713.92 N

So, the magnitude of the force that stops the bullet is 