caitybugking
04.07.2020 •
Physics
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-the loop which has a radius 5 m. The acceleration of gravity is 9.8 m/s^2.
Required:
How large is the normal force on it if its mass is 5g?
Solved
Show answers
More tips
- A Animals and plants How to Grow Lime from a Seed: Simple Tips and Interesting Facts...
- C Computers and Internet How to Create a Folder on Your iPhone?...
- G Goods and services How to sew a ribbon: Tips for beginners...
- F Food and Cooking How to Make Mayonnaise at Home? Secrets of Homemade Mayonnaise...
- C Computers and Internet Which Phone is Best for Internet Surfing?...
- F Food and Cooking Everything You Need to Know About Pasta...
- C Computers and Internet How to Choose a Monitor?...
- H Horoscopes, Magic, Divination Where Did Tarot Cards Come From?...
- S Style and Beauty How to Make Your Lips Fuller? Ideas and Tips for Beautiful Lips...
Answers on questions: Physics
- P Physics Mary walked north from her home to sheila s home, which is 4.0 kilometers away. then she turned right and walked another 3.0 kilometers to the supermarket, which is...
- M Mathematics Find the volume of the composite figure....
- H History Andrew jackson and his supporters believed in a) the necessity of the national bank b) limited government control of the economy c) government control of the economy...
Ответ:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down
Ответ:
The answer is C: nervous, defensive
Explanation: