Answer:
Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck
So, the acceleration of the hockey puck is 
.
No one expected violet & ultraviolet spectral lines to be shifted towards the red.
The ball is against the vector of gravity. Then, the gravity will be negative.
The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.
It will stay approx. 9.44 seconds in the air.
Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
The area of the top and bottom:
2πr²
Cost for top and bottom:
2πr² x 0.02
= 0.04πr²
Area for side:
2πrh
Cost for side:
2πrh x 0.01
= 0.02πrh
Total cost:
C = 0.04πr² + 0.02πrh
We know that the volume of the can is:
V = πr²h
h = 500/πr²
Substituting this into the cost equation to get a cost function of radius:
C(r) = 0.04πr² + 0.02πr(500/πr²)
C(r) = 0.04πr² + 10/r
Now, we differentiate with respect to r and equate to 0 to obtain the minimum value:
0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π
r = 3.41 cm
