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Lunna [17]
3 years ago
14

The Petronas twin towers in Malaysia and the Chicago Sears tower have heights of about 452 m and 443 m respectively. If objects

were dropped from the top of each, what would be the difference in the time it takes for the objects to reach the ground?
Physics
1 answer:
fredd [130]3 years ago
8 0

Answer:

0.09 s

Explanation:

From the second equation of motion,

S=ut+\frac{1}{2}at^{2}

Here, u is the initial velocity, a is the acceleration due to gravity, t is time taken, and S is the total displacement or distance.

From the given problem,

initial velocity is zero for both the case.

And the distance of twin  tower of malaysia is, S_{1}=452m

And the distance of Sears tower of Chicago is, S_{1}=443m

Now,rearrange the distance equation for t.

t=\sqrt{\frac{2S}{a} }

So time difference.

\Delta t=\sqrt{\frac{2(452)}{9.8} }-\sqrt{\frac{2(443)}{9.8} }\\\Delta t=\sqrt{92.244898}-\sqrt{90.4081633}  \\\Delta t=9.60 s-9.51s\\\Delta t=0.09 s

Therefore, the difference in time, object will reach the ground is 0.09 s

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WINSTONCH [101]

Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

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