Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
The relationship between wavelength
, frequency f and speed of light c for an electromagnetic wave is
Using the data of the problem, we find
Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
Answer:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:
where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is
Re-arranging for r
And substituting into the previous equation
Solving for T,
So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
In the process of peppering the question with those forty (40 !) un-necessary quotation marks, you neglected to actually show us the illustration. So we have no information to describe the adjacent positions, and we're not able to come up with any answer to the question.
