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DIA [1.3K]
3 years ago
5

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

Physics
2 answers:
olasank [31]3 years ago
8 0

There could be more than just one answer, since kilograms can be converted to grams, to miligrams, etc.

7.2\frac{g.m}{s}

or

7200\frac{mg.m}{s}

Why?

Let's remember some conversion factors to work with kilograms (kg)

1kg=1000g=1x10^{3}g \\1kg=1,000,000mg=1x10^{6}mg \\1g=1000mg=1x10^{3}mg

So, we are given the momentum:

momentum=7.2x10^{-3} \frac{kg.m}{s}

We can rewrite the units of the momentum (equivalent) as follow:

momentum=7.2x10^{-3}kg \frac{m}{s}=7.2x10^{3} x10^{-3}g\frac{m}{s}=7.2\frac{g.m}{s}

and

momentum=7.2x10^{-3}kg \frac{m}{s}=7.2x10^{3} x10^{6}mg\frac{m}{s}=7.2x10^{3}\frac{mg.m}{s}

Note: There could be more equivalent units for the momentum, in example, we could work with equivalent units for meters (distance) and seconds (time).

Have a nice day!

pishuonlain [190]3 years ago
7 0

Answer:

7.2*10^{-3} kg*\frac{m}{s}= 7.2 g* \frac{m}{s}

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a  ⇒ b

c ⇒ x

Then

x=\frac{c*b}{a}

The direct rule of three is the rule applied in this case to perform the change of units.

The rule of three applies as follows: being 1 kg equivalent to 1000 g, then 7.2*10⁻³ kg m / s, how many grams will they be (keeping m / s without unit changes)?

\frac{7.2*10^{-3} kg*\frac{m}{s} *1000 g}{1 kg}

7.2 g*\frac{m}{s}

Then <u><em>7.2*10^{-3} kg*\frac{m}{s}= 7.2 g* \frac{m}{s}</em></u>

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
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Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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Answer:

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