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Brilliant_brown [7]
2 years ago
15

Can anybody write a short poem about friction

Physics
1 answer:
Luden [163]2 years ago
8 0
Answer:

you will be the clouds
and I will be the sky.
you will be the ocean
and I will be the shore.
you will be the trees
and I will be the wind.

whatever we are, you and I will always collide.

There you go! Let me know if it helped.
:)
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A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
4 years ago
A worker pushes horizontally on a 35.0 kg crate with a force of magnitude 112 N. The coefficient of static friction between the
OLga [1]

Answer: a) -127 N b) No. c) -112 N  d) 40 N e) 15 N

Explanation:

a) Friction force always oppose to the relative movement between two surfaces, and, provided that be less than the fs max, adopt any value to counteract the applied force.

The fs max, is the horizontal component of the contact force, and can be written as follows:

Fs max = us . Fn  

As the block is at rest in the vertical direction, this means that Fn must be numerically equal to the weight of the object:

Fn = m g = 35 kg. 9.8 m/s2 = 343 N → Fs max = 0.37. 343 N = 127 N

b) Now, as the applied force is smaller than Fs max, this means that the friction force, is equal and opposite to the applied forcé, i.e., -112 N, so the crate doesn´t move.

c) Please see above.

d) As explained above, the maximum friction force, is proportional to the normal force, which adopts any value needed to satisfy the Newton´s 2nd Law.

So, if we diminish the normal force, we can lower the máximum friction force, helping to the worker to move the crate.

The mínimum needed normal force, will be the one that satisfies the following:

Fs max = F applied = us Fn = 112 N = 0.37. Fn.

Solving for Fn, we get Fn= 303 N

So, the difference between the original normal force and the new one, will be the mínimum upward force needed to make the crate to move, as follows:

Fup = 343 N -303 N = 40 N  

e) If we keep the normal force unchanged, but add an horizontal force to help the worker, we will need that the sum of both forces, will be equal to the Fs max, as follows:

Fh + Fapp = 127 N → Fh = 127 N – 112 N = 15 N

5 0
3 years ago
Identify forces between two objects which are inversely proportional to the square of the distance between them. i. nuclear forc
nikdorinn [45]
The force between two celestial bodies is a Newtonian gravitational force. 
It is also called Newton's law of universal gravitation. We can write it down mathematically in the following way:
F=G \frac{ m_{1} m_{2} }{ r^{2}}
We can see from this formula that gravitational force is <span>inversely proportional to the square of the distance between bodies.
The electrostatic force between two charges is Coulombs force. We can write it down like this: 
</span>F= k_{e} \frac{ q_{1} q_{2} }{ r^{2}}
This force is also inversely proportional to the square of the distance between interacting particles.
The nuclear force is a little bit more complicated. It can be expressed using the so-called Yukawa potential, which has the following form:
V_{y} =- g^{2} \frac{ e^{-ur} }{r}
This interaction does not follow the inverse-square law.
The final answer should be 8.
3 0
3 years ago
Arnold Schwarzenegger is moving out of his California mansion. He pushes a box of books on the floor with a force of 50 N to the
sleet_krkn [62]
A.draw in the forces .Use the scale 1cm =10n of force 
3 0
4 years ago
Oil having a density of 924 kg/m 3 floats on water. A rectangular block of wood 3.79 cm high and with a density of 970 kg/m3 flo
Katena32 [7]

Answer:

x = 2.69 cm

Explanation:

We are given ;

Density of oil; ρ_oil = 924 kg/m³

Density of wood; ρ_wood = 970 kg/m³

h = 3.79cm

Density of water ( ρ_water) has a constant value of 1000 kg/m³

At equilibrium position, we have;

ρ_wood•g•h - ρ_oil•g•(h - x) - ρ_water•g•x = 0

This is because the density of oil is lower than that of water while density of wood is higher than that of oil but lower than that of water.

x is the distance below the interface between the two liquids is the bottom of the block.

Thus, let's make x the subject;

x = [(ρ_wood - ρ_oil)/(ρ_water - ρ_oil)] x h

Plugging in the relevant values to get ;

x = [(970 - 924)/(1000 - 924)] x 3.79

x = (54/76) x 3.79 = 2.69cm

5 0
4 years ago
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