Answer:
The answer is B complimentary
Explanation:
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
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Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo
Answer:
Explanation:
When the spring is compressed by .80 m , restoring force by spring on block
= 130 x .80
= 104 N , acting away from wall
External force = 82 N , acting towards wall
Force of friction acting towards wall = μmg
= .4 x 4 x 9.8
= 15.68 N
Net force away from wall
= 104 -15.68 - 82
= 6.32 N
Acceleration
= 6.32 / 4
= 1.58 m / s²
It will be away from wall
Energy released by compressed spring = 1/2 k x²
= .5 x 130 x .8²
= 41.6 J
Energy lost in friction
= μmg x .8
= .4 x 4 x 9.8 x .8
= 12.544 J
Energy available to block
= 41.6 - 12.544 J
= 29 J
Kinetic energy of block = 29
1/2 x 4 x v² = 29
v = 3.8 m / s
This will b speed of block as soon as spring relaxes. (x = 0 )
