Answer:
a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV
Explanation:
a. KE =1/2 (MV^2) where the M is mass of electron
b. E = V/d
c. V= 0 V (momentarily the pd changes to zero)
d KE= 300*3600 v = 1.08 MeV
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Since they are in the same direction, you would add them together. Let’s also assume said direction is positive. 225 N + 165 N = 390 N
Answer:
a) The speed is 61.42 m/s
b) The drag force is 10.32 N
Explanation:
a) The Reynold´s number for the model and prototype is:


Equaling both Reynold's number:

Clearing Vm:

b) The drag force is:

The anwser is C it can also be A but C would make the most sense