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saw5 [17]
3 years ago
11

What prevents friction

Physics
2 answers:
zheka24 [161]3 years ago
7 0
A smooth and slick surface?:)
GuDViN [60]3 years ago
5 0

Answer:

control of friction can be done with the help of lubricants

Explanation:

You might be interested in
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
3 years ago
A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?
Digiron [165]

Answer:

1962

Explanation:

w = f × d

= 20×9.81 × 10

= 1962

5 0
3 years ago
What does atomic spectrum mean?
lapo4ka [179]

Answer:

The range of characteristic frequencies of electromagnetic radiation that are readily absorbed and emitted by an atom. The atomic spectrum is an effect of the quantized orbits of electrons around the atom

Additional Facts:

  • Atomic spectra can also be analyzed to determine the composition of objects
  • The frequency depends on the difference in energy between the orbitals. Explaining this phenomenon was crucial to the development of quantum mechanics
  • Occurs due to the fact are quantized at specific levels determined by the atomic number

3 0
3 years ago
A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
4 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
4 years ago
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