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quester [9]
3 years ago
12

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

an is only 0.400 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)? N
Physics
1 answer:
IRINA_888 [86]3 years ago
5 0

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = \frac{Frictional\ force}{coefficient\ of\ friction}        ............1

put here value and we will get value

normal force = \frac{0.400}{0.04}  

solve it we get

normal force = 10 N

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When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials A) cond
Alexandra [31]

When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.

Answer: Option C

<u>Explanation:</u>

We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.

But for some materials, like when we rubbed a balloon with human hair, we observed clouding of excess static charge on the balloon surface. This indicates that hair can easily give up electrons to balloon leading to clouding of excess static charge on it.

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A boat is rowed perpendicular to the shore of a river that flows at 3.0 m/s as shown in the
vazorg [7]
The magnitude is 5 ms^-1

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Two bullets of the same size, mass and horizontal velocity are fired at identical blocks, only one is made of steel and the othe
tatiyna

Answer and Explanation:

  • Since we're discussing shots, the significant thing is the way the energy is changed over as there is deceleration of the bullet to a halt when it hits something.
  • Kinetic Energy is relative to mass times speed squared, so in reality, the 2 cases given have practically indistinguishable Kinetic energy. The measure of energy is authoritative, so the two cases will do generally a similar harm given, obviously we look at situations when all the kinetic energy is spent.
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  • Henceforth, the heavier yet more slow bullet with a similar energy will venture to every part of a similar separation in the engrossing material, but since of bigger force, will take a more drawn out time doing it.
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Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45
victus00 [196]
<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>
4 0
3 years ago
In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro
blondinia [14]
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
                                        plussitive on the proton

Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

                  A  =  (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !
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