Answer:
The minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)
mv²/r = mg
v² = gr = 9.8 × 25 = 245
v = 15.65 m/s
But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop
Change in kinetic energy = potential energy at the top
Change in kinetic energy = (mv₂² - mv₁²)/2
v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s
v₂ = minimum velocity the car must have at the top of the loop to not fall
And potential energy at the top of the loop = mgh (where h = the diameter of the loop)
(mv₂² - mv₁²)/2 = mgh
(v₂² - v₁²) = 2gh
(v₂² - (15.65)²) = 2×9.8×50
v₂² - 245 = 980
v₂² = 1225
v₂ = 35 m/s
Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s
Answer:
1 m = 100 cm....so 2.5 m = (2.5 * 100) = 250 cm
a = 1st shelf
b = 2nd
c = 3rd
d = 4th
a + b + c + d = 250
b = 2a + 18
c = a - 12
d = a + 4
a + (2a + 18) + (a - 12) + (a + 4) = 250
5a + 10 = 250
5a = 250 - 10
5a = 240
a = 240/5
a = 48 cm <== 1st shelf
b = 2a + 18 = 2(48) + 18 = 114 cm <== 2nd shelf
c = a - 12 = 48 - 12 = 36 cm <== 3rd shelf
d = a + 4 = 48 + 4 = 52 cm <== 4th shelf
so 2nd shelf is 114 cm
Protons- 17
neutrons-20
electrons-17
<u>Answer</u>:
Effort is the unaltered force. Load is the altered force.
Answer:
125.66 R/s
Explanation:
First 1200 r / min = 20 r/sec
20 r/s * 2pi Radians / r = 40 pi Radians / sec = 125.66 R/s
