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astraxan [27]
3 years ago
6

Sobre un cuerpo de 700 g de masa que se apoya en una mesa horizontal se aplica una fuerza de 5N en la dirección del plano. Calcu

la la fuerza de rozamiento si:
a)El cuerpo adquiere una aceleración igual a 1,5 m/s2

b)El cuerpo se mueve con velocidad constante.
Physics
1 answer:
kicyunya [14]3 years ago
6 0

Answer:

(a) Ff = 3.95N

(b) Ff = 5N

Explanation:

(a) To find the friction force you use the second Newton law:

F=ma

But you take into account the applied force and the opposite friction force:

F_a-F_f=ma

you do Ff the subject of the formula and replace the values of the parameters:

F_f=F_a-ma=5N-(0.700kg)(1.5m/s^2)\\\\F_f=3.95N

(b) In the case of a motion with constant velocity you have that there is no acceleration. Hence:

F_a-F_f=0\\\\F_f=F_a\\\\F_f=5N

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graph would be a straight line from (0, 0) to (400, 8)

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2 years ago
Is friction used when buttering a cake pan?
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A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?
MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

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Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

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The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
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