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11111nata11111 [884]
4 years ago
7

A block is moving at constant speed due to a horizontal force pulling to the right. The coefficient of kinetic friction, Hk, bet

ween the block and the surface is 0.20 and the mag- nitude of the frictional force is 100.0 N, what is the weight of the block? (a 400 N (b) 600 N (c) 500 N (d) 267 N

Physics
1 answer:
VashaNatasha [74]4 years ago
4 0

Answer:

c) 500 N

Explanation:

If the block moves with constant speed there is no acceleration. We draw a free body diagram and define the forces on the body.

F_{f}=FrictionForce\\F_{e}=ExternalForce\\N=NormalForce\\W=Weight\\

The equation for the frictional force is:

F_{f}=N*H_{k}

Where H_{k} is the kinetic friction coefficient. We write the equilibrium equation in the y direcction:

\sum F_{y}\rightarrow N-W=0\\N=W

We replace this result in the equation for the frictional force:

F_{f}=W*H_{k}

We replace the data given by the exercise and find the weight

W=\frac{F_{f}}{H_{k}}\\W=\frac{100}{0.2}=500\: N

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98.1 Joule

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Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

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In this case we have the surface of a sole of a tennis shoe:

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On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

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Substituting (2) and (3) in (1):

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What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
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K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

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From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

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We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

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