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babymother [125]
3 years ago
10

Which one is the dependent variable in distance, force, or work

Physics
1 answer:
Elina [12.6K]3 years ago
3 0

Answer:

Distance

Explanation:

Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.

Mathematically, work done is given by the formula;

W = F * d

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

A dependent variable is the event expected to change when an independent variable is manipulated.

Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.

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The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

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An automobile tire having a temperature of −1.6 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 22 lb/in2 . What
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Answer:

25.8 lb/in²

Explanation:

Gay-Lussac's law tells us that given an ideal gas of a certain mass has a constant volume, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} } \\\\\frac{22}{-1.6+273.15} =\frac{P_{2} }{45+273.15} \\\\P_{2} = \frac{22*318.15}{271.55} = 25.8lb/in^{2}

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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
Sav [38]

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}.

7 0
3 years ago
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