Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
Answer:
0.076 m/s
Explanation:
Momentum is conserved:
m v = (m + M) V
(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V
V = 0.076 m/s
After catching the puck, the goalie slides at 0.076 m/s.
Answer:
The work done by gravity is ![4.975 \: Joules](https://tex.z-dn.net/?f=4.975%20%5C%3A%20Joules)
Explanation:
The data given in the question is :
Mass is ![0.245 kg](https://tex.z-dn.net/?f=0.245%20kg)
Height from ground is ![2.07 m](https://tex.z-dn.net/?f=2.07%20m)
As we know , the work done is state function , it depends on initial and final position not on the path followed.
So, work done by gravity = change in potential energy
Work done = Initial potential energy - final potential energy
Insert values from question
Work done = ![mass \times gravity \times (change \: in \: height)](https://tex.z-dn.net/?f=mass%20%5Ctimes%20gravity%20%5Ctimes%20%28change%20%5C%3A%20in%20%5C%3A%20height%29)
Work done = ![0.245 kg \times 9.81 m/s^{2} \times 2.07 m](https://tex.z-dn.net/?f=0.245%20kg%20%5Ctimes%209.81%20m%2Fs%5E%7B2%7D%20%5Ctimes%202.07%20m)
So, work done = ![4.975 Joules](https://tex.z-dn.net/?f=4.975%20Joules)
Hence the work done by gravity is ![4.975 \: Joules](https://tex.z-dn.net/?f=4.975%20%5C%3A%20Joules)
Answer: work Melvin did=9000J
Explanation:
Given to complete the question: If the sled moved 33.9m,how much work did Melvin do? Answer in unit of J and round to the nearest thousandth.
W = F ×S
W = 317 × cos 33°×33.9
W=9012.6055J
W=9000J to the nearest thousandth
Answer:
The critical radius of the plastic insulation is 0.72 inches.
Explanation:
Given that,
Diameter = 0.091 in
Thickness = 0.02 in
Initial temperature = 90°F
Final temperature = 50°F
Heat transfer coefficient = 2.5 Btu/h.ft²°F
Material conductivity = 0.075 Btu/h.ft °F
We need to calculate the critical radius of the plastic insulation
Using formula of critical radius
![r_{cr}=\dfrac{2K}{h}](https://tex.z-dn.net/?f=r_%7Bcr%7D%3D%5Cdfrac%7B2K%7D%7Bh%7D)
Where, k = Material conductivity
h = Heat transfer coefficient
Put the value into the formula
![r_{cr}=\dfrac{2\times0.075}{2.5}](https://tex.z-dn.net/?f=r_%7Bcr%7D%3D%5Cdfrac%7B2%5Ctimes0.075%7D%7B2.5%7D)
![r_{cr}=0.06\ ft](https://tex.z-dn.net/?f=r_%7Bcr%7D%3D0.06%5C%20ft)
![r_{cr}=0.72\ inches](https://tex.z-dn.net/?f=r_%7Bcr%7D%3D0.72%5C%20inches)
Hence, The critical radius of the plastic insulation is 0.72 inches.