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Tomtit [17]
3 years ago
8

How does static electricity apply to car paint application?

Physics
1 answer:
Talja [164]3 years ago
4 0

Answer:

For electrostatic paint to be applied, the vehicle needs to be grounded and positively charged. This creates a magnetic attraction to the negatively charged paint. Due to this charge, when the paint leaves the nozzle, it is attracted to the vehicle's charge and will stick to it.

Explanation:

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A rock is stuck deep in the dirt. When you pull on the rock to remove it, what type of force are you exerting?
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It is Tension as the other 3 answer choices would not make sense. Compression would mean you are pressing the rock on both sides or in this case, pushing it into the dirt. It can't be nuclear force as you are pulling out a rock. Air resistance would not make sense either as there is no air involved in the scenario at all.
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Which phenomena did Michael Faraday explore to help him invent the motor and the generator?
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To Build The Motor, Faraday Studied Electricity And Magnetism. Hope This helps!
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An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How w
Ksivusya [100]

It will be approximately equal.

<h3>How will the final kinetic energy change?</h3>

We can infer that all of the energy in the electron is Potential energy (PE) because the energy provided by the photon is hardly enough to outweigh the work function.

It will gain kinetic energy (KE) as it advances in the direction of the anode because it is moving through an electric field. All of the PE will have been transformed to KE by the time it reaches the anode.

According to the question

K = hf - W

W = Work function

The energy of photons is comparable. After conversion, there was only a little amount of KE remaining.

Therefore, PE (W) essentially equals KE (K).

It will about be equal.

Learn more about work function here:

brainly.com/question/19595244

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3 0
2 years ago
A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the
Tema [17]

The cat fell 1.0 m from the ground.

Using the formula

h = v_{oy}t - \frac{gt^2}{2}

Here, v_{oy} = 0, \; h = 1.0 \; m.

Solving for t, the time it spent in the air is

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s


5 0
3 years ago
The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start
Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction \mu _s=0.24

Coefficient of kinetic friction \mu _k=0.22

mass of crate m=20\ kg

Force applied F=200\ N

maximum static Friction F_s=\mu _sN

N=mg

F_s=0.24\times 20\times 9.8

F_s=47.04\ N

thus applied force is greater than Static friction therefore kinetic friction will come into play

F_k=\mu _kN

F_k=0.22\times 20\times 9.8=43.12\ N

net Force on crate F-F_k=ma

a=\frac{200-43.12}{20}=7.84\ m/s^2

Magnitude of velocity can be obtained by using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

v^2-0=2\times 7.84\times 2

v=\sqrt{31.37}

v=5.60\ m/s                                          

7 0
4 years ago
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