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3 years ago

What is the major difference between the natural frequency and the damped frequency of oscillation.

Physics

1 answer:

natita [175]3 years ago

6 0

Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

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A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m = 0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m $U^{2}$

K.E = 1/2 x 0.17 x $6^{2}$

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/ $s^{2}$

To calculate the final velocity, let us use third equation of motion.

$V^{2}$ = $U^{2}$ + 2as

$V^{2}$ = $6^{2}$ + 2 x 0.3 x 61

$V^{2}$ = 36 + 36

$V^{2}$ = 72

V = $\sqrt{72}$

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

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Each time the heart beats,what does it do to the blood

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3 years ago

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Answer: c

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3 years ago

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To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
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csc θ = 1 / sin θ

Using these identities:
cot θ ∙ sec θ = (cos θ / sin θ ) ( 1 / cos θ)

We can cancel out cos θ, leaving us with
cot θ ∙ sec θ = 1 / sin θ
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3 years ago

a 5-kg fish swimming at 1 m/s swallows an absent minded 1-kg fish at rest. What is the speed of the large fish immediately afyer

storchak [24]

Answer:

In first case speed of fish will be $\frac{5}{6}$ m/s.

In the second case the speed of fish $\frac{1}{6}$ m/s

Explanation:

Given data :-

Mass of bigger fish ( m₁ ) = 5 kg.

Mass of small fish ( m₂ ) = 1 kg.

Speed of large fish ( v₁ ) = 1 m/s

Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.

Case - 1

Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s

Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v

v = speed of bigger fish immediately after lunch.

Using the conservation of momentum.

m₁* v₁ = ( m₁ + m₂)*v

5 = 6 * v

v = $\frac{5}{6}$ m/s.

Case -2

Speed of small fish = 4 m/s

Momentum of bigger fish before lunch = 5 kg.m/s

Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s

Net momentum before lunch = 5 - 4 = 1 kg.m/s

Momentum of bigger fish after eating the larger fish = 6 * V

Using the conservation of momentum.

1 = 6 * V

V = $\frac{1}{6}$ m/s.

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3 years ago

What is the major difference between the natural frequency and the damped frequency of oscillation.​

What is the major difference between the natural frequency and the damped frequency of oscillation.