Cause surface currents to move in circular paths.
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:
give brainliest please
Explanation:
T- lymphocytes or T cells
Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=
