Answer:
correct option is a. 0.2 mA toward D
Explanation:
given data
B carries = 1.5 mA
C carries current = 1.3 mA
solution
we take positive direction of current going away from the point D
and negative direction of current coming towards point D
so we use here kirchoff's current law
that is
iA + iB + iC = 0 ......................1
iA + 1.5 + (-1.3) = 0
iA = - 0.2 mA
so that current in wire A is 0.2 mA towards point D
correct option is a. 0.2 mA toward D
Sliding and Static.
Would be the right one here.
Answer:
a)0.024
b)0.148
Explanation:
Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H
Given:
P(L) = 0.16
P(H) = 0.10
P(L n H) = 0.1 ·P( L u H )
Hence, P( L u H) = 10 ·P( L nH)
(a)
Hence. using the equation. P(L U H) = P(L) + P(H) - P(L n H)
Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )
Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26
Hence, P(L n H) =
0.26/11=0.024
(b)
We know that condition probability P(H ║ L) = p(L n H)/P(L)
hence, P(H ║ L) =(0.26/11)/0.16 =0.148
Atoms are made up or one or more molecules
Answer:
0.06
Explanation:
On a level surface, normal force = weight.
Sum of the forces in the x direction:
∑F = ma
Nμ = ma
mgμ = ma
gμ = a
μ = a/g
μ = (0.6 m/s²) / (9.8 m/s²)
μ = 0.06
