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NARA [144]
3 years ago
8

A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for

ward along the floor, gradually slowing as it moves. Consider the horizontal force(s) on the cart while it is moving forward and slowing. Which of the following statements is correct?
Question 36 options:

Both a forward and a backward force are acting on the cart, but the forward force is larger.

Only a forward force is acting which diminishes with time.

Both a forward and a backward force are acting on the cart, but the backward force is larger.

Only a backward force is acting, no forward force.
Physics
1 answer:
mash [69]3 years ago
8 0

Answer:

Only a backward force is acting, no forward force.

Explanation:

  • Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
  • As it would move at a constant speed, no net force would be acting on it.
  • So, if it is gradually slowing, there must  be a net force producing an acceleration in a direction opposite to the movement.
  • This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
  • As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
  • This is consistent with the direction of the acceleration of the cart.
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Nature of components of mixture should be known to separate the mixture why?​
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Explanation:

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3 years ago
Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is
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Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

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8 0
3 years ago
A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
  • Mass of the box, m = 3.1 kg
  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
  • Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

Learn more about work done here: brainly.com/question/8119756

8 0
2 years ago
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