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Marrrta [24]
3 years ago
11

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
Physics
1 answer:
Lana71 [14]3 years ago
5 0

Answer:

a

   m  = 0.169 \ kg

b

  |v_{max} |=  0.5653 \ m/s

Explanation:

From the question we are told that

    The  spring constant is  k =  14 \ N/m

     The  maximum extension of the spring is  A =  6.0 \ cm  =  0.06 \ m

     The number of oscillation is  n  =  30

      The  time taken is  t  =  20 \ s

Generally the the angular speed of this oscillations is mathematically represented as

           w = \frac{2 \pi}{T}

where T is the period which is mathematically represented as

     T  =  \frac{t}{n}

substituting values

     T  =  \frac{20}{30 }

     T  = 0.667 \ s

Thus  

       w = \frac{2 * 3.142 }{ 0.667}

       w =  9.421 \ rad/s

this angular speed can also be represented mathematically as

       w =  \sqrt{\frac{k}{m} }

=>   m  =\frac{k }{w^2}

substituting values

      m  =\frac{ 15 }{(9.421)^2}

      m  = 0.169 \ kg

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        v =  - Awsin (wt)

The  velocity is maximum when  wt = \(90^o) \ or \ 1.5708\ rad

     v_{max} = -  A* w

=>   |v_{max} |=  A* w

=>    |v_{max} |=   0.06 * 9.421

=>   |v_{max} |=  0.5653 \ m/s

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A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length
Ivahew [28]

Answer:

<em>The total length of the spring would be 0.65 m</em>

Explanation:

The Concept

Hooke's law evaluates the increment of  spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the  force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;

F = k x ..................1

where F is the force applied to the spring

k is the spring constant

x is the spring stretch or extension

Step by Step Calculations

We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;

x = F/k

but F = m x g

so, x = (m x g)/k

given that, the mass of the person m =150 kg

g is the acceleration due to gravity = 9.81 m/s^{2}

k is the spring constant = 10000 N/m

then x = (9.81 m/s^{2} x 150 kg)/10000 N/m

x = 0.14715 m

the extension experienced by the spring after the compression is 0.14715 m

The total length of the spring would be;

L = 0.14715 m + 0.5 m = 0.64715

L ≈  0.65 m

Therefore the total length of the spring would be 0.65 m

4 0
3 years ago
If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveli
NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

  • When two waves with different frequency travelling in the same medium meet each other, they produce an interference pattern called beat.
  • <em><u>The frequency of the beat produced is equivalent to </u></em><em><u>the difference between the individual frequencies of the two waves involved.</u></em>
  • <em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
  • We need to find a pair from the choices whose frequency difference is 20 Hz.
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8 0
3 years ago
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aliya0001 [1]

a^2+b^2=c^2

20km^2+13km^2=c^2

400km^2+169km^2=c^2

23.853720...km

7 0
3 years ago
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How much tension must a rope withstand if used to accelerate a 960-kg car from rest horizontally along a frictionless surface to
Svetach [21]
Use a=(dv/dt)         (change in velocity/ change in time)=acceleration
(1.2/5)=acceleration

F=ma (Newton's second law, Force= Mass x Acceleration

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7 0
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Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

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