By using Ohm's law, we can find what should be the resistance of the wire, R:
Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
So the area is
And by using the resistivity of the Aluminum,
, we can use the relationship between resistance R and resistivity:
to find L, the length of the wire:
<u>Answer</u>:
The coefficient of static friction between the tires and the road is 1.987
<u>Explanation</u>:
<u>Given</u>:
Radius of the track, r = 516 m
Tangential Acceleration 
= 3.89 m/s^2
Speed,v = 32.8 m/s
<u>To Find:</u>
The coefficient of static friction between the tires and the road = ?
<u>Solution</u>:
The radial Acceleration is given by,
Now the total acceleration is
=>
=>
=>
=>
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is
From (1) and (2)
Substituting the values, we get
Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
If she only has 21 buttons and all 21 of them are large, then all of her buttons are large. so 100% of the buttons would be large.
