Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
Answer: 405.3 minutes
Explanation: In order to explain this problem we have to use the following:
Fisrtly we calculate the volume of the wire, this is given by:
Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.
Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3
then the number of the total electrons in tthe wire volume is given by;
n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23
Finally, considering the current in the wire equal to 4.4*10^18 electrons/s
the time consuming to extract all the electrons from the wire is given by:
t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s
equivalent to 405.3 minutes
Answer:
a. The angular frequency is doubled.
e. The period is reduced to one-half of what it was.
Explanation:
Angular frequency is given as;
ω = 2πf

when the frequency is doubled

Thus, the angular frequency will be doubled.
Amplitude in simple harmonic motion is the maximum displacement.
Frequency is related to period in simple harmonic motion as given in the equation below;

when the frequency is doubled;

Thus, the period will be reduced to one-half of what it was.
Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa