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tankabanditka [31]
2 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]2 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]2 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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Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

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Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

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1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

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Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

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