Explanation:
It is given that,
Length of the copper wire, l = 4.4 m
Diameter of copper wire, d = 1.3 mm = 0.0013 m
Radius of copper wire, r = 0.00065 m
The resistivity of the copper wire, 
We need to find the resistance of the copper wire. It is given by :
R =0.055 ohms
So, the resistance of the copper wire is 0.055 ohms. Hence, this is the required solution.
Sodium (Na) is very reactive because it does not have a full valence shell.
Answer:
Impulse = change in momentum w bounce
There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:
- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:
- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:
- Replacing by the givens in (5), we can solve for μsmín, as follows:
Answer:
if i'm not mistaken that's either a plug in cord for a certain device, or what i am assuming to be a usb software cord.
Explanation:
