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3 years ago

A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour. What is

Physics

1 answer:

madam [21]3 years ago

6 0

<h2>How to solve?</h2>

We know that, Velocity is the rate of displacement covered. Displacement is the shortest path between the Initial and Final point covered by the body. So,

Velocity = Displacement / Time

And, when it comes to Average velocity, It is the total displacement by total time taken. So, By using this let's solve this question.....

<h2>Solution:</h2>

✏️ Refer to the attachment...

Let the body goes to point A that is 7 m East of the Initial point. Then it comes backward because West is opposite to East in perpendicular direction. It covers 1.5 m backwards in the same line to reach B which is the Final point.

So,

Displacement = Final point - Initial point

⇛ Displacement = 7 m - 1.5 m

⇛ Displacement = 5.5 m

Total time taken,

⇛ 2 hours + 1 hour

⇛ 3 hours

Finding Average displacement,

⇛ Total displacement / Total time taken

⇛ 5.5 m / 3 hours

⇛ 1.83333.... hours

So, the Final answer is,

$\huge{ \boxed{ \bold{ \purple{1.8 \bar{3}m}}}}$

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4 years ago

Calculate the value of ∆g for these initial partial pressures. (the partial pressure of each gas is set a 5. 0 atm at 25°c) (use

Sati [7]

∆g for these initial partial pressures is 10,403.31 KJ.

ΔG gets increasingly positive as a product gas's partial pressure is raised. ΔG becomes more negative as the partial pressure of a reactant gas increases.

∆g = RT ln (q/k)

In this equation: R = 8.314 J mol⁻¹ K⁻¹ or 0.008314 kJ mol⁻¹ K⁻¹

K = 325

If ΔG < 0, then K > Q, and the reaction must proceed to the right to reach equilibrium.

∴∆g = RT ln (q/k)

= 8.314 × 298 ln ( 5 / 325)

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= 2477.57 × (-4.199)

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Products are preferred over reactants at equilibrium if G° 0 and both the products and reactants are in their standard states. When reactants are preferred above products in equilibrium, however, if G° > 0, K 1. At equilibrium, neither reactants nor products are preferred if G° = 0, hence K = 1.

Therefore, ∆g for these initial partial pressures is 10,403.31 KJ.

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1 year ago

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correct answer is Hudson bay

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