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4 years ago

What happens to the coefficient of friction when the weight is increased? Why is this?

Physics

2 answers:

sp2606 [1]4 years ago

8 0

Answer:

The coefficient of friction does not change.

Explanation:

The reason for this is that the coefficient friction depends only on one variable, which is the nature of the surfaces in contact. Therefore, weight has no effect whatsoever on the coefficient of friction. hence, No change occurs in the coefficient of friction if the weight is changed.

hope it helped ya.

Crazy boy [7]4 years ago

6 0

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

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Photoelectric effect:

Lynna [10]

Answer:

A. K = 0.546 eV

B. cooper and iron will not emit electrons

Explanation:

A. This is a problem about photoelectric effect. Then you have the following equation:

$K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi$ (1)

K: kinetic energy of the ejected electron

Ф: Work function of the metal = 2.48eV

h: Planck constant = 4.136*10^{-15} eV.s

λ: wavelength of light = 410nm - 750nm

c: speed of light = 3*10^8 m/s

As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :

$K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV$

B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm

$E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV$

You compare the energies E1 and E2 with the work functions of the metals and you can conclude:

sodium = 2.3eV < E1

cesium = 2.1 eV < E1

cooper = 4.7eV > E1 (this metal will not emit electrons)

iron = 4.5eV > E1 (this metal will not emit electrons)

5 0

4 years ago

Which type of wave does the illustration depict?

NeX [460]

Answer is B. Longitudinal Wave

5 0

4 years ago

Read 2 more answers

Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

amid [387]

Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

8 0

3 years ago

If you lose control of your vehicle and collide with a fixed object, such as a tree, at 60 m.p.h., the force of impact is the sa

My name is Ann [436]

You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.

First, convert 60 mph to m/s:

60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s

Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.

free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)

H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m

If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.

Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.

7 0

4 years ago

If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

dsp73

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.

7 0

3 years ago

Read 2 more answers