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
The skittles because they are small so you can fit more in
1.25 is the answer to tht if im right but could be wrong
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A
Answer:
15.6m/s
Explanation:
V1=
because the derivate of the position is the velocity
V1=12t+3
V2=20+
-8t because the integral of the acceleration is the velocity
V2=
V1=V2 to see when the velocities of particles match
12t+3=20-4t^2
4t^2+12t-17=0 we resolve this with 
and we take the positif root
t=1.05 sec
if we evaluate the velocity (V1 or V2) the result is 15.6m/s
