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nydimaria [60]
3 years ago
13

The straight line in the experiment of hooks law does not pass through the origin beacause?

Physics
1 answer:
pshichka [43]3 years ago
7 0
Okay, let’s look at it this way: when does a line pass through the origin? The line represents possible values of x and y that satisfy the equation.

So, when a line passes through the origin, it passes through the coordinates (0,0). x = 0, y = 0. So, let’s model this with the equation ax + by + c = 0. Sub in x = 0 and y = 0 to the equation and we find that 0 + 0 + c = 0. Clearly, c = 0.

So, with this simple explanation, I hope you understand when a line does pass through the origin.

Now, let’s look at when a line doesn’t pass through the origin. This is when c is not equal to 0. Hence, when x = 0, y cannot equal 0; c + by = 0, and we know that c is not 0. If y is 0, then we get c = 0… where c is not 0. Ehh. Thus, you can see that a line does not pass through the origin when c is not equal to 0 by ehat is hope is a simple explanation. You don’t need to know how to prove it, I presume, but that’s not too hard either.

Oops, I realised that I just assumed you were talking about linear graphs. For quadratic graphs, the reasoning is similar. For graphs of the form y = ax^2, the minimum/maximum point of the graph will be the origin. For graphs of the form y = ax^2 + bx, it will pass through the origin but the line of symmetry will be different. For graphs of the form y = ax^2 + bx + c (you know, where c is not zero) , the graph will not pass through the origin because the maximum/ minimum point is actually raised or lowered by c units
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
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Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
J.J. Thomson discovered the electron by noticing that: A. molecules with the same atoms exhibited the same chemical properties.
GenaCL600 [577]

J.J. Thomson discovered the electron by noticing that a beam of particles could be influenced by an electric or magnetic force.. That is option B.

<h3>What is an electron?</h3>

An electron can be defined as the part of an atom that is negatively charged and is found revolving round the nucleus of an atom.

J.J. Thomson was the scientist that discovered electrons through subjecting two oppositely-charged electric plates around the cathode ray.

He noticed that the particles where deflected by both the magnetic and electric fields.

Learn more about cathode rays here:

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4 0
2 years ago
HOW DO PROTONS, NEUTRONS AND ELECTRONS DIFFER?
aliina [53]
Protons are positively charged and neutrons are neutral whereas electrons are negatively charged.
6 0
3 years ago
Read 2 more answers
It takes 160 kj of work to accelerate a car from 24.0 m/s to 27.5 m/s. what is the car's mass?
ankoles [38]
Work-Energy :W = 1/2 m ( Vf^2 -Vo^2 )
Vo = 24.0 m/s Initial speed 
 Vf = 27.5 m/s  Final speed 

W = 1/2 m ( Vf^2 -Vo^2 )
160 kj = 1/ 2 m ( 27.5^2  -24.0 ^2)
160kj =  4680 x m
convert kilo joules to jeoules                     160000 j = 4689 xm
m = 160000 j/4689
m = 34.18 kg
4 0
3 years ago
A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca
sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09  m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

d  = ut + \frac{1}{2}at^2

u = 0 starting from rest

d = \frac{1}{2}at^2

t^2 = \frac{2d}{a}

for bike

d+25 = 0 + \frac{1}{2}*4.40t^2

t^2= \frac{d+25}{2.20}

equating time of both

\frac{2d}{a} = \frac{d+25}{2.20}

solving for d we get

d = 132 m

therefore t is= \sqrt{\frac{2d}{a}}

t =  \sqrt{\frac{2*132}{3.70}}

t = 8.45 sec

each travelled in time 8.45 sec as

for car

d = \frac{1}{2}*3.70 *8.45^2

d = 132.09  m

fro bike

d = \frac{1}{2}*4.40 *8.45^2

d = 157.08 m

7 0
3 years ago
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