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Physics Static Question

Physics static equilibrium question?

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Physics static electricity question?

It depends if you are including the effects of relativity or not. If you want to be accurate, the speed is fast enough for relativity to make a significant difference.
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Energy, E, gained by a charge q passing through V volts is:
E = qV. = 1.6x10^-19 x 30,000 = 4.8 x 10^-15J.
In this case E will equal the kinetic energy (1/2)mv^2

To get the speed, first assume relativistic corrections are not needed,
(1/2)mv^2 = E
v = sqrt(2E/m) = sqrt(2 x 4.8x10^-15 / 9.11x10^-31) = 1.03x10^8 m/s

Since this is about 1/3 of the speed of light, the answer is not accurate because of relativistic effect. For a more accurate answer we use:
Total energy = mc2/g, where g = sqrt(1-(v/c)^2)
('g' should really be a Greek letter 'gamma')

KE = Total energy - rest mass energy = mc^2/g - mc^2
= mc^2((1/g) - 1)
= mc^2((1-g)/g)
KE/mc^2 = (1-g)/g

Use A= KE/mc^2 for brevity.
Note A = 4.8 x 10^-15 / (9.11x10^-31 x (3x10^8)^2) = 0.05854

A = ((1-g)/g)
Ag = 1-g
g(A+1) = 1
g = 1/(A+1) = 1/1.05854 = 0.9447

sqrt(1-(v/c)^2) = 0.9447
1-(v/c)^2 = 0.892
v/c = sqrt( 1- 0.892) = 0.329
v = 0.329c = 9.86 x 10^7 m/s
This is a bit less than the answer ignoring relativity.

Physics: Static friction question?

An 78.2 N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 21.1° to the ground.
(a) What is the normal force exerted on the crate by the ramp?
(b) What is the static frictional force exerted on the crate by the ramp?
(c) What is the minimum possible value of the coefficient of static friction?
(d) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. Find the magnitude and direction of the contact force.

Try to push a large desk with some force say 10 N. Let us say the desk does not move. In this case your force is being neutralized by a force of friction that is static in nature (because the desk is static or stationery) and equal to 10N, acting in the opposite direction.Now let us say you increase the force to 12 N, and the desk still does not move. It is not moving since the static force of friction has now increased itself to 12 N to oppose the force you are impressing on the desk.You can keep increasing this force and the static force of friction will also increase to the same value as you are applying to ensure that the desk does not move.However, a time will come when the static friction will no more be able to surpass the force that you are putting on the desk. This is therefore called the limiting value of static friction and is often calculated as Fs = μs R. Where Fs is the limiting force of friction, μs is the static coefficient of friction and R is the normal reaction (on a flat table it will be mg)So you see the static force of friction is not constant. It increase with the applied force and reaches a maximum value that is calculated by the above formula.Learn more in this video about the concept -How Static and Dynamic Friction fits into Newton's Laws of Motion #7

Physics questions about static electricity?

Most modern applications of electricity involve moving electric charges or current electricity. Historically, however, the first studies of electricity involved static charges, or electrostatics. In this course we follow history, starting with electrostatics, and this experiment is an exercise in electrostatics.

The simplest way to obtain an electric charge is to take almost any two materials and rub them together. It is found that one of the materials becomes negatively charged and the other positively. For example, if you rub a hard rubber rod with a woolen cloth, the rubber rod acquires a negative charge and the wool an equal amount of positive charge. (Which charge we call negative is a matter of convention, but with the agreed convention it is the rubber that becomes negative.)

We now know that all matter contains positive and negative charges. The positive charge is deep inside each atom on the atomic nucleus, while the negative charge is on the electrons that orbit outside the nucleus. In normal, neutral matter the amounts of positive and negative charge are equal, and we are unaware of either. Given any two materials, it usually happens that one has a greater attraction, or affinity, for electrons than the other; when the two are rubbed together, the material with higher affinity captures a few electrons from the material with lower affinity. For example, rubber has higher electron affinity than wool; when they are rubbed together the rubber captures a few electrons from the wool; the rubber therefore acquires an excess of electrons and is negatively charged, while the wool is left with a deficit of electrons and is therefore positively charged.

Physics question : mobile, static equilibrium?

Just start balancing your torques for each section of the mobile starting at the bottom. Having a net torque of 0 produces a state of rotational equilibrium meaning its gonna stay horizontal.
The equation for calculating torque is:
t = r F sin(Angle)
where:
t, is the torque
r, is the distance from pivot point the force is being applied
F, is the magnitude of the force being applied
Angle, the angle at which the force is being applied and since in this case case always going to be 90 degrees and sin(90) = 1 the equation can be simplified to just:
t = r F

Start at the bottom
rotational equilibrium like i said before is achieved when the net t is zero. so:
t net = 0
but the tnet is being provided by 2 torques the 125g weight on the left (t1) and the unknown one on the right (t2). so:
t1 - t2 = 0
F1 r1 - F2 r2 = 0
since the only force in this case is being provided by the weight of the weights then your F = mg
m1g r1 - m2g r2 = 0
m1g r1 = m2g r2
m1 r1 = m2 r2
m2 = [ m1 r1 ]/[ r2 ]
now looking at this equation you know everything except for the value of m2 plug in and solve for it.
m2 = [ (.125kg)(.3m) ]/[ .2m ]
m2 = .1875 kg = 187.5g
That would be the mass of the weight label A.
The rest of the problem is just a repetition of this except that at the next layer the force acting at the 10cm side of the pivot is going to be the combine weight of the 2 bottom weights ( 125g + 187.5g = 312.5g ) and at the level above that its going to be the combines weight of the bottom 3 weights.
After doing the math you should get
weightC = 250g
weightB = 104.2g
weightA = 187.5g

Where's the rod's center of mass? What are the vertical and horizontal forces? How great a (static) coefficient of friction is required to overcome the horizontal force at the point of contact?

As described, it is at the end of it’s journey and stopped. Therefore speed is zero.

Physics - Static and Kinetic Friction?

Hi guys, just a quick question. I'm stuck on these three (well two) questions:
.
1) 2 movers push a 260kg box across the tiles. John pushes with 280N[forward] and Adrian pushes with 340N[forward]. The box accelerates with 0.30m/s^2. How long will it take for the box to stop moving after pushing it 6.2s from rest? (Note: the coefficient of kinetic friction is 0.21)
.
2) a 65kg runner accelerates from rest into a breeze. The breeze has a force of 65N opposite of the runner's direction. The earth exerts a constant forward force of 250N on the runner's feet.
. Calculate the sprinter's acceleration.
. Calculate the distance travelled in the first 2.0s
. Calculate the minimun coefficient of friction between the runner's shoes and the ground.
. Explain, is the friction applied on the runner from the ground static or kinetic?
.
Thank you, that's all my questions for now. Please include ALL formulas, and work done to achieve the answer. I know the answers, just not how to get to them, please and thank you!!

With the developing technology and the ambitions of humankind, physics is always likely to discover questions it doesn't have answers to yet. This particular longing for finding answers and exploring new dimensions which led to historical discoveries. Take the Higgs Boson particle for example.Among all the talk and scares , when the Large hadron collider went operational, it was a feat of enormous proportions, both literally and figuratively. And the existence of Higgs Boson particle was discovered and standard model was confirmed. This particle is so unique in the way it interacts with other particles, it's amazing. This also explains why some particles do have mass when they should be mass-less. It's kind of an origin story to the particles family of Quantum physics and may be, the whole Universe. And this, leads to further questions and ventures in this area. Coming to present day questions, the behavior and characteristics of Higgs Boson particles is a massive question as it may explain a lot about the Universe. Cool Origin story, don't you think?Another that comes to mind is Black holes. Not much is known about them. Scientists ponder and ponder but are clueless about it being a massive paradox. While it's said the singularity prevents anything escaping a black hole, they still emit radiation. It's as if the black hole is saying "Go figure".

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