Calculate the tangential speed of a yo-yo twirled at the end of a 4 meter long string at 2 revolutions per second.

Physics

1 answer:

Oliga [24]2 years ago

8 0

Answer: 50 m/s

Explanation: speed v = 2· pi·n·r = 2· 3.14· 2 s^-1· 4 m

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A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock l

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A geologist is studying rock layers in an old river bed, and he finds a fossil of a fish and a horsetail rush in the same rock layer. According to the law of faunal and floral succession, the geologist can assume that the rock containing the fossils may date back as far as the <span>Devonian period</span>.

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The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

$Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%$

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

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Why is it important to use democratic method of choosing someone for a position of authority

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So that the correct representative of the people gets elected to head the people- so that a person who is thought to be capable enough to rule over people and take the correct decisions may be representative to the people.

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A hot iron ball of mass 200 g is cooled to a temperature of 22°C from 100°C. How much heat was

Nookie1986 [14]

Answer:

Q= -6900 J

Explanation:

use the formula Q=mC(T_2 - T_1) and sub in givens

Q=mC(T_2 - T_1)

Q= (200 g)(0.444 J/g°C)(22-100)

Q= -6900 J