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Alex
3 years ago
9

A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

oscillating system is 6.74 J.
A. Determine the following maximum speed of the object?

B. The force constant?

C. The amplitude of the Motion?
Physics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

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Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

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3 years ago
1. A 4-N force is used to move an object 2 m in 10 s. Which of the following is the power generated while moving the object?
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Power is the work done per unit time. Therefore,

\begin{gathered} p=\frac{w}{t} \\ \text{where} \\ w=\text{work done} \\ t=\text{time} \end{gathered}

Therefore,

\begin{gathered} \text{work done=fd} \\ f=force=4N \\ d=\text{displacement}=2m \\ t=2\sec s \\ p=\frac{fd}{t}=\frac{4\times2}{10}=\frac{8}{10}=0.8watts \end{gathered}

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1 year ago
The ice skaters partner liftes her up a distance of 1 m work done or not work done
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Answer:

Work done.

Explanation:

The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.

(This all assumes lifting vertically and no other forces being part of the picture)

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3 years ago
Explain relative velocity.
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Answer:

The relative velocity of an object A with respect to another object B.

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Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions? ​
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Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

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