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tankabanditka [31]
3 years ago
5

The drawings show three examples of the force with which someone pushes against a vertical wall. In each case the magnitude of t

he pushing force is the same. Rank the normal forces that the wall applies to the pusher in ascending order (smallest first).

Physics
2 answers:
zaharov [31]3 years ago
7 0

Answer:

drawing 2< drawing 1 < drawing 3

Explanation:

Complete Question is:

The drawings(attachment) show three examples of the force with which someone pushes against a vertical wall. In each case the magnitude of the pushing force is the same. Rank the normal forces that the wall applies to the pusher in ascending order (smallest first)

1st drawing is a medium angle

2nd drawing is a small angle

3rd drawing is a large angle (almost 90 degrees and perpendicular to the wall)

Normal Force= Fsinθ

Here is θ is the angle between the wall and the force applied that is F.

As θ increases, value of sinθ increases. When θ is 90, sinθ is maximum that is 1 and value of resultant.

Therefore, normal force is greatest in diagram 3 and least in diagram 2

Monica [59]3 years ago
4 0

Complete Question

The complete question is shown on the

Answer:

The ascending order would be 2nd < 1st < 3rd

Explanation:

Generally the the Normal force is mathematically represented as

                      Normal \ Force  = Fsin \theta

=>                    Normal \ Force \  \alpha  \  sin \theta

For the first drawing  the value   \theta is between that of the the second and the third drawing so the Normal force would also be between the normal forces of the second and the third drawing

For the second drawing whose value of \theta is less than that of the first and the third the normal force would also be less than that of the first and third

For the the third drawing whose value is (90°) which is higher than the first and the second the normal force would also be higher than the first and the second

         

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What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
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3 years ago
If a car with a mass of 1,200 kg traveling westward at 30 m/s is
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The net force acting on the car is equal to -12000 Newton.

<u>Given the following data:</u>

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  • Initial velocity = 30 m/s.
  • Final velocity = 0 m/s (since the car came to a stop).
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To determine the net force acting on the car:

<h3>How to calculate the net force.</h3>

Mathematically, the net force acting on the car is given by Newton's Second Law of Motion:

F = \frac{m(v\;-\;u)}{t}

<u>Where:</u>

  • m is the mass.
  • t is the time.
  • u is the initial velocity.
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Substituting the given parameters into the formula, we have;

F = \frac{1200 \times (0\;-\;30)}{3}\\\\F = \frac{1200 \times (-30)}{3}\\\\F = \frac{-36000 }{3}

Net force = -12000 Newton.

Read more on net force here: brainly.com/question/1121817

5 0
2 years ago
How do I find the tension? its supposed to be in the mid 40s​
babunello [35]

Answer:

a = 2.3 m/s²

T = 45 N

Explanation:

Draw a free body diagram for each mass.

For the mass on the incline, there are four forces:

Weight force mg pulling down.

Normal force N perpendicular to the incline.

Friction force Nμ pushing down the incline.

Tension force T pulling up the incline.

For the hanging mass, there are two forces:

Weight force Mg pulling down.

Tension force T pulling up.

Sum of the forces on the hanging mass in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of the forces on the sliding mass in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces on the sliding mass in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Substitute:

Mg − Ma − mg sin θ − mgμ cos θ = ma

Mg − mg (sin θ + μ cos θ) = ma + Ma

Mg − mg (sin θ + μ cos θ) = (m + M) a

a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)

Plug in values:

a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)

a = 2.3 m/s²

Now find tension:

T = Mg − Ma

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5 0
4 years ago
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