Assume that an ingot of copper has a mass of 9.1 kg or 9100 g.
The cross-sectional area of the copper wire with diameter of 6.5 mm (or 0.65 cm) is
A = (π/4)*(0.65 cm)² = 0.3318 cm²
The density of copper is given as 8.94 g/cm³.
If the length of copper wire is L cm, then
(0.3318 cm²)*(L cm)*(8.94 g/cm³) = 9100 g
L = 9100/(0.3318*8.94) = 3.0678 x 10³ cm
Note that
1 cm = 1/2.54 in = 1/2.54 in = 0.3937 in
= 0.3937/12 = 0.03281 ft
Therefore
L = (3.0678 x 10³ cm)*(0.03281 ft/cm) = 100.65 ft
Answer: 100.65 ft
The part that causes the disc caliper piston to retract when the brakes are released is the square-cut O-ring.
The square cut seal is the most important part of a disc brake caliper, for keeping the brake behind the piston so that when you step on the brake pedal, it releases a pressure that applied to the piston which in return applies the pad to the rotor.
Answer:
![0.44 m/s^2](https://tex.z-dn.net/?f=0.44%20m%2Fs%5E2)
Explanation:
We have the following data:
- distance covered by the child: d = 2 m (length of the slide)
- time taken to cover this distance: t = 3 s
- initial velocity of the child: 0 m/s (he starts from rest)
So we can find the acceleration by using the equation:
![d=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=d%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
Where a is the acceleration.
Substituting the values and solving for a,
![a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2d%7D%7Bt%5E2%7D%3D%5Cfrac%7B2%282%29%7D%7B3%5E2%7D%3D0.44%20m%2Fs%5E2)
Lower. Water expands on lower temperatures, meaning less molecules in 1 m3, thus making it less dense