the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
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:
Explanation:
D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m
ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s
We can apply the equation
Ff = W ⇒ μ*N = m*g <em>(I)</em>
then we have
N = Fc = m*ac = m*(ω²*R)
Returning to the equation <em>I</em>
<em />
μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)
Finally
μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379
Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
