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scZoUnD [109]
3 years ago
11

1. What is the arrangement of the outer planets? 2. What effect does their placement have the planets?

Physics
1 answer:
Rasek [7]3 years ago
4 0
<span>the arrangement of the outer planets is 
</span>1. Mercury 
<span>2. Venus </span>
<span>3. Earth </span>
<span>4. Mars </span>
<span>5. Jupiter </span>
<span>6. Saturn </span>
<span>7. Uranus </span>
8. Neptune
the inner most of the outer plannets is jupitor it is followed by saturn uranus and neptune
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A large block attached to a spring is undergoing simple harmonic motion horizontally, with angular frequency ω and amplitude A.
Naily [24]

Answer:

Explanation:

It is a problem based on simple harmonic motion and friction . maximum friction possible between large block and mass m is µ mg . During SHM , maximum acceleration is ω² A and force is  mω² A .

friction must exceed it so that mass m may not slip over it during motion . so

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4 years ago
An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

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