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JulsSmile [24]
3 years ago
10

A sports car is advertised to be able to stop in a distance of 50.0 m from a speed is 99.0km/h. What is its acceleration in m/s2

?
Physics
1 answer:
Shtirlitz [24]3 years ago
4 0
99.0km/h =27.5m/s (this is the initial speed)
The final speed is zero
The distance is 50.0m
Therefore you use the formula:
vfinal²=vinitial²+2ad
a=(vfinal²-vinitial²)/2d
 = (0²-27.5²)/(2x50.0)
 =-7.5625 or in correct sigdigs -7.56m/s²
Hope this helps!
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dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

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Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L =  90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s            (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s               (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁²  (inside)      =       P₂ + pgh₂ + 1/2PV₂²   (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

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P₂ = 236500 Pa

cheers I hope this helped !!

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Answer:

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