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AleksAgata [21]
3 years ago
13

According to the graph above, how large of a force is needed in order to stretch the string 1.00 meters?

Physics
2 answers:
gulaghasi [49]3 years ago
7 0

My guess for this one would be; 400 N

My reasoning would be; it starts at 0 on both X and Y, if you need to get to 1.00 meters thats 4/4. 1/4 of 1.00 is .25, and on .25 its on 100 so multiply it by 4 to make 1.00 and you get 400 N

MrMuchimi3 years ago
6 0

Answer:

The needed force is 400 N.

Explanation:

We need to calculate the force for 1.00 m string

According to graph,

The force applied for stretch from 0.10 to 0.15 is 20 N.

0.10-0.15 = 0.05 m

So, the force applied is 20 N for 0.05 m.

For stretch 0.05 m = 20 N

For stretch 1 m = \dfrac{20}{0.05}= 400\ N

Hence, The needed force is 400 N.

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Answer:

I=\frac{4}{3}ML^2+2MR^2+2MRL

Explanation:

We are given that

Mass of rod=M

Length of rod=L

Mass of hoop=M

Radius of hoop=R

We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.

Moment of inertia of rod about center of mass=\frac{1}{12}ML^2

Moment of inertia of hoop about center of mass=MR^2

Moment of inertia of the pendulum about the pivot left end,I=\frac{1}{12}ML^2+M(\frac{L}{2})^2+MR^2+M(L+R)^2

Moment of inertia of the pendulum about the pivot left end,I=\frac{1}{12}ML^2+\frac{1}{4}ML^2+MR^2+MR^2+ML^2+2MRL

Moment of inertia of the pendulum about the pivot left end,I=\frac{1+3+12}{12}ML^2+2MR^2+2MLR

Moment of inertia  of the pendulum about the pivot left end,I=\frac{16}{12}ML^2+2MR^2+2MRL

Moment of inertia of the pendulum about the pivot left end,I=\frac{4}{3}ML^2+2MR^2+2MRL

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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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