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expeople1 [14]
4 years ago
12

A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6

5, and the coefficient of kinetic friction is 0.35. What horizontal force must be exerted on the box for it to accelerate at 1.2 m/s2?
A) 60 N
B) 230 N
C) 120 N
D) 170 N
E) 490 N
Physics
1 answer:
dsp734 years ago
6 0

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

\sum F = ma

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

F - F_f = ma

To find the required force then,

F=F_f+ma

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

F = \mu mg +ma

F = 0.35*50*0.8+50*1.2

F=(171.5N)+(50Kg)(1.2m/s^2)

F=231.5N

F\approx 230N

Therefore the horizontal force applied on the block is B) 230N

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Hope this helps!
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3 years ago
A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
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<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

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Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

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<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
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x1 = 14 ft + 1 ft = 15 ft

<span>
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3 0
4 years ago
An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one nega
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Explanation:

It is given that,

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F=\dfrac{kq^2}{r^2}

r=\sqrt{\dfrac{kq^2}{F}}

r=\sqrt{\dfrac{9\times 10^9\times (1)^2}{650}}

r = 3721.04 meters

So, the distance between the humans is 3721.04 meters. Hence, this is the required solution.

5 0
3 years ago
A 83 kg man is resting on the roof top of his house. If he has a potential energy of
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60 m because that’s the top energy of the roof
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3 years ago
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The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b
bagirrra123 [75]

To solve this problem it is necessary to address the concepts related to Torque as a function of the force and distance where it is applied and the moment of inertia from which the torque, moment of inertia and angular acceleration are related.

By definition the torque is defined as

\tau = F*r

Where,

\tau = Torque

F = Force

r = Radius

For our values we have:

\tau = F*r

\tau = (1.75*10^3)(2.8*10^{-2})

\tau = 49Nm

Consequently the calculation of the moment of inertia would then be given by the relationship

\tau = I\alpha

I=\frac{\tau}{\alpha}

Replacing with our values

I = \frac{49}{150}

I = 0.322Kg.m^2

The moment of inertia of the boxer's forearm 0.322Kg.m^2

4 0
3 years ago
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