I’m assuming you’re supposed to calculate the resultant force?
425N (right) -300N (left)
=125 N to the right
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
Answer:
It means zero acceleration.
Explanation:
When a body moves with constant velocity then it is said to be there is no rate of change of velocity takes place because it is constant. So the acceleration becomes zero or uniform acceleration.
Answer:
The coefficient of kinetic friction = 0.026
Explanation:
An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.
Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.
The external force acting on the body is in the forward direction and the friction acts in the backward direction.
Friction increases as the mass of the body increases.
Friction = ![u_{k}\times m \times g](https://tex.z-dn.net/?f=u_%7Bk%7D%5Ctimes%20m%20%5Ctimes%20g)
We now equate this to the external force of 15 N.
15 = ![u_{k} \times 56 \times 10](https://tex.z-dn.net/?f=u_%7Bk%7D%20%5Ctimes%2056%20%5Ctimes%2010)
= ![\frac{15}{560}](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B560%7D)
= 0.026
The coefficient of kinetic friction = 0.026