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3 years ago

15

you can use a torque feeler that consists of a meterstick held at the 0 cm end with a weight dangling from various positions alo

ng the stick. When the stick is held horizontally, torque produced when a 1 kg object hangs fromt he 50 cm mark. How much torque is exerted when the object is hung from the 75 cm mark

1 answer:

Deffense [45]3 years ago

7 0

Answer:

$T=7.3575\ N.m$

Explanation:

Given:

length of the stick, $l=100\ cm=1\ m$

position of holding the stick that acts as a hinge, $f=0\ cm$

mass of the object, $m=1\ kg$

position of hanging the mass on the meter stick, $x=50\ cm=0.5\ m$

new position of hanging the mass, $x'=75\ cm=0.75\ m$

<u>Now the torque in the initial position:</u>

$T=mg.x$

where:

$g=$ acceleration due to gravity

$T=1\times 9.81\times 0.5$

$T=4.905\ N.m$

Now when the mass is at the new position:

$T=mg.x'$

$T=1\times 9.81\times 0.75$

$T=7.3575\ N.m$

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

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3 years ago

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4 years ago

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Energy in the form of motion is potential energy.<br> True<br> False

Veseljchak [2.6K]

False

Energy in the form of motion is kinetic energy

Stored energy is called potential energy

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3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

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To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

$u = 0m/s$

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

$s = ut+\frac{1}{2} at^2$

Replacing,

$1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2$

$u =6.4013m/s$

The height of the acorn above the meter stick can be calculated as,

$v^2 = u^2 +2gh$

$h = \frac{v^2-u^2}{2g}$

$h = \frac{6.4013^2-0^2}{2(9.8)}$

$h = 2.0906m$

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

$h = 2.0906+1.87$

$h = 3.9606m$

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4 years ago