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klemol [59]
3 years ago
13

A bead slides without friction around a loopthe-loop. The bead is released from a height 18.6 m from the bottom of the loop-the-

loop which has a radius 6 m. The acceleration of gravity is 9.8 m/s 2 . 18.6 m 6 m A What is its speed at point A ? Answer in units of m/s.
Physics
1 answer:
12345 [234]3 years ago
3 0

Answer:

See explanation

Explanation:

To do this, you need to use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path so, you can write the expression like this:

 

1/2mv1² + mgh1 = (1/2)mv2² + mgh2

Where:

v1 = 0 because it's released from rest

h1 = 18.6 m

v2 = speed we want to solve.

h2 = height at point A. In this case, you are not providing the picture or data, so, I'm going to suppose a theorical data to solve this. Let's say h2 it's 12 m.

Now, let's replace the data in the above expression (assuming h2 = 12 m). Also, remember that we don't have the mass of the bead, but we don't need it to solve it, because it's simplified by the equation, therefore the final expression is:

1/2v1² + gh1 = 1/2v2² + gh2  

Replacing the data we have:

1/2*(0) + 9.8*18.6 = 1/2v² + 9.8*12

182.28 = 117.6 + 1/2v²

182.28 - 117.6 = v²/2

64.68 * 2 = v²

v = √129.36

v = 11.37 m/s

Now, remember that you are not providing the picture to see exactly the value of height at point A. With that picture, just replace the value in this procedure, and you'll get an accurate result.

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8 0
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A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator la
gladu [14]

Answer:

<h2>a) 496N</h2><h2>b) 50.56kg</h2><h2>c) 1.90m/s²</h2>

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

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Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = -ma

N2 = -ma+Fg ...(2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

<em>\b) To get the person mass, we will use the relationship Fg = mg</em>

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

3 0
4 years ago
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