If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Answer:
Time, t = 0.015 seconds.
Explanation:
Given the following data;
Mass, m = 0.2kg
Force, F = 200N
Initial velocity, u = 40m/s
Final velocity, v = 25m/s
To find the time;
Ft = m(v - u)
Time, t = m(v - u)/f
Substituting into the equation, we have;
Time, t = 0.2(25 - 40)/200
Time, t = 0.2(-15)/200
Time, t = 3/200
Time, t = 0.015 seconds.
Note: We ignored the negative sign because time can't be negative.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A
Explanation:
Em = 223 V
f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF
a)
Capacitive reactance = Xc=?
Xc= 
Xc=1/2pi *399*23.1*10^-6
Xc= 17.267 Ω
b).
Z=
Xl= 2π * f * L
Xl= 2π * 399 * 147 * 
Xl= 368.5 Ω
Z= = 
Z= 415.5 Ω
c).
Current:
I= V / Z= Em / Z
I= 223/415.5
I= 0.537 A
