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Travka [436]
3 years ago
11

The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f

loor) is 98.2 kPa.
Determine the height of the building if the air density is assumed to be independent of the height, being 1.16 kg/m³, while the acceleration of gravity is 9.81 N/kg.
Physics
1 answer:
Stells [14]3 years ago
6 0

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

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Answer:

Part a)

v_a = 3.94 m/s

Part b)

v_b = 3.35 m/s

Part C)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

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a = -\frac{F_f}{m}

a = -\frac{\mu mg}{m}

a = - \mu g

now we will have

v_f^2 - v_i^2 = 2ad

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v_a = \sqrt{(2(0.13)(9.81)(6.1)}

v_a = 3.94 m/s

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

v_b = \sqrt{2(0.13)(9.81)(4.4)}

v_b = 3.35 m/s

Part C)

By momentum conservation we will have

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

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Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

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Mathematically this frequency is described as

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Replacing we have,

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