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Pani-rosa [81]
3 years ago
8

Ethan pushes a wooden box across a carpeted floor. Then he pushes the same box across a smooth marble floor. Why does Ethan find

it easier to push the box on the marble floor than on the carpeted floor?
A. The box experience more friction on the carpeted floor
B. The box experience less friction on the carpet floor
C. The box experiences more applied force on the marble floor
D. The box experiences less normal force on the marble floor
Physics
2 answers:
FinnZ [79.3K]3 years ago
4 0
Answer A is correct because the carpet usually enforces more friction. The marble carpet is smoother, thus there is less friction
AysviL [449]3 years ago
3 0

Answer:

A. The box experiences more friction on the carpeted floor

Explanation:

Friction is the force that opposes the motion of an object when it slides along a surface. The magnitude of the friction is given by

F=\mu mg

where m is the mass of the object, g is the acceleration due to gravity, and \mu is the coefficient of friction, which depends on the type of material of the surface: the larger this coefficient, the stronger the friction, the more difficult is to push the box along the surface. Generally, a smooth surface has a lower coefficient of friction, while a rough surface has a larger coefficient of friction.

In this case, Ethan find it easier to push the box on the marble floor, because marble is smoother than the carpet and so friction is weaker, while for the carpeted floor the coefficient of friction is larger and so the friction is stronger, making it more difficult to push the box.

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Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

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x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

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Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
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Answer:

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Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

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to find out

material work function

solution

we know by Einstein photo electric equation  that is

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h ( c / λr ) = Ф +  kinetic energy

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now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

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put all value

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Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

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