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Hunter-Best [27]
3 years ago
12

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (

Note: 1 kg = 1000 g).
Physics
1 answer:
slavikrds [6]3 years ago
6 0

Answer:

Momentum, p = 7.2 g-m/s

Explanation:

It is given that,

The momentum of an object is p=7.2\times 10^{-3}\ kg-m/s

We need to express momentum in any equivalent units. There can be many solutions of this problem. Some of the units of mass are gram, milligram etc. units of length are meters, mm etc.

Since, 1 kg = 1000 gram

So, p=7.2\times 10^{-3}\times 10^3\ g-m/s

Therefore, the momentum of the object is 7.2 g-m/s. Hence, this is the required solution.

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A block with a mass M = 4.85 kg is resting on a slide that has a curved surface. There is no friction. The speed of the block af
natali 33 [55]

Answer:

The correct option is a

Explanation:

From the question we are told that

   The mass of the block is  m =  4.84 \ kg

    The height of the vertical  drop is h =  19.6 \  m

Generally from the law of energy conservation , the potential energy at the top  of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as

        PE  =  KE

i.e     m *  g  *   h   =  \frac{1}{2} *  m *  v^2

=>    gh  =  0.5 v^2

=>   v = \sqrt{\frac{9.8 *  19.6}{0.5 } }

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4 0
3 years ago
Walt ran 5 km in 25 minutes going east to what was his average velocity
goldenfox [79]

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time taken = 25 minute = 25 x 60 sec = 1500 sec

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V = 5000/1500

V = 3.33 m/s towards east

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a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t
laiz [17]

Answer: A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

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Cross multiply

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V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

4 0
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