Answer:
a) The student feel light
b) Nbottom = 758 N
c) N'top= 236 N
d) N'bottom= 1055 N
Explanation:
a) W= 659N , Ntop= 560N
W > Ntop ---> Student feel less weight
b) Top:
∑F= W - Ntop = m.v²/R
m.v²/R = 659N - 560 N = 99 N
Bottom:
∑F= Nbottom- W = m.v²/R
Nbottom= W + m.v²/R = 659N + 99 N = 758N
c) W= 659 N , Ntop= 560 N , v'=2.v
N'top= ?
∑F= W - N'top = m.v'²/R
N'top= W - 4.m.v²/R
N'top = 659 N - 4. 99 N = 263 N
d) N'bottom = ?
∑Fbottom= N'bottom- W = m.v'²/R
N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N
d. the surface on which the object is moving is made smoother
This is called lubrication, I think ...
b. the weight of the moving object is decreased
If there is less araea of contact because of the weight reduction ("squeezing together") then this could also have an effexct. I thinl though the thrust of the q is on lubrication = oil, talcum powder, thin films of water etc
Please put answers in the comments of actual answers, as I need them! Lots of points! 63 points
The mutualism I believe. So sorry if I’m wrong
R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%
Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
