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Leto [7]
3 years ago
14

2. A 14.8-kg child sits in a 1.30-kg swing. You pull the swing back, lifting it 52.1 cm vertically, and then let go. Determine t

he speed of the child and swing as the swing moves past its lowest point.
Physics
1 answer:
drek231 [11]3 years ago
6 0

To answer this problem, we have to define the two energies, Potential and Kinetic Energy.

Initially when the child is at rest at 52.1 cm, the potential energy PE is at its highest.

PE = m g h

Where,

m = total mass = 14.8 kg + 1.30 kg = 16.10 kg

g = gravitational acceleration = 9.8 m /s^2

<span> h = height = 52.1 cm = 0.521 m</span>

Computing for PE:

PE = 16.10 * 9.8 * 0.521

PE = 82.29 N

As the swing moves past its lowest point h = 0, then all PE is converted to Kinetic Energy KE. Therefore, at lowest point

KE = 82.29

KE = 0.5 m v^2

0.5 (16.10) v^2 = 82.29

<span>v = 3.20 m/s</span>

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If I did my math right I believe 13 is the 9.3 seconds and 14 is the first option because of logic.  The question basically says, how does the height change and the answer is the first one because it gets higher then lower, or increases then decreases.

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3 years ago
An initially electrically neutral conducting sphere is placed on an insulating stand. A negatively-charged glass rod is brought
balu736 [363]

Answer: Option (c) is the correct answer

Explanation:

It is known that when a neutral object comes in contact with a charged object then an opposite charge develops on the neutral object. But this development of charge occurs only when both the objects come in contact and if they are not in contact with each other then there occurs no charge on the neutral object.

As in the given situation, negatively-charged glass rod is brought near, but does not touch the sphere. So, there will occur no charge on the sphere.

Thus, we can conclude that final charge on the sphere is neutral.

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3 years ago
When a person's temperature reaches 104 degrees, the chances of survival are decreased dramatically. Group of answer choices Tru
Natali5045456 [20]

Answer:

True

Explanation:

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4 years ago
Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as prot
lubasha [3.4K]

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

          F = q v x B

bold letters indicate vectors, the magnitude of this expression is

          F = q v B sin θ  

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

           a = v² / r

let's use Newton's second law

           F = ma

           q v B = m v² / r

           r = \frac{qB}{mv}

Let's apply this expression to our case.

Proton 1

             r = \frac{qB_1}{mv_1}

Proton 2

             r₂ = \frac{q \ B_2}{m \  v_2}

in the exercise indicate some relationships between the two protons

*    v₁ = 2 v₂

    v₂ = v₁ / 2

*   B₂ = 2B₁

we substitute

           r₂ = \frac{q \ 2B_1}{m \ \frac{v_1}{2} }

           r₂ = 4 \frac{qB_1}{mv_1}

           r₂ = 4 r

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