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Len [333]
3 years ago
12

A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin stri

ng is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

linear acceleration

a = \frac{2g}{2 + \frac{R}{b}}

angular acceleration

\alpha = \frac{2g}{R(2 + \frac{R}{b})}

Explanation:

As we know that the force due to tension force is upwards while weight of the disc is downwards

so we will have

2mg - T = 2ma

also we have

Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha

now we have

Tb = mR^2(\frac{a}{R})

T = \frac{mRa}{b}

now we have

2mg = (2ma + \frac{mRa}{b})

a(2 + \frac{R}{b}) = 2g

so we have

linear acceleration

a = \frac{2g}{2 + \frac{R}{b}}

angular acceleration

\alpha = \frac{2g}{R(2 + \frac{R}{b})}

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An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it
KATRIN_1 [288]

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

The ball reaches a height of 2.8 m

We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

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3 years ago
The electric current leaves the battery through the --- complete this blank
aev [14]
The bulb

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5 0
3 years ago
Give the relationship(s) for any pair of protons with the proper term(s). Label – your choice. A.Heterotopic B.Heterotopic, dias
Afina-wow [57]

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

7 0
3 years ago
Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air
Aleonysh [2.5K]

Answer:

a) I = 0 N s,  b)  v = -3.935 m / s, c) vf = 3.935 m / s,   d)   y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

          I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

        I = ∫ (9200 t - 11500 t2) dt

        I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

       I = 9200 (0.8² -0) - 11500 (0.8³ -0)

       I = 5888 -5888

       I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

      v² = v₀² - 2g y

     v = √ (0 - 2 9.8 (-0.79))

     v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

      I = ΔP

      I = m vf - m v₀

     vf = (I + m v₀) / m

This is the impulse of women on the floor    

      vf = ( 0 + 68 (3.935)) / 68

      vf = 3.935 m / s

d) let's use kinematics

      v₂ = v₀² - 2gy

      0 = v₀² - 2gy

      y = v₀² / 2g

      y = 3.935²/2 9.8

      y = 0.790 m

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3 years ago
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