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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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3.8 m/s
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Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

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We know that from Newton's Second Law;

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F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
2 years ago
An integrated circuit is a ___.
EastWind [94]
HI buddy!

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5 0
2 years ago
Help please :pensive:
tino4ka555 [31]

Answer:

0m/s²

Explanation:

Given parameters:

Initial velocity of the boat = 8m/s

Final velocity  = 8m/s

Time taken  = 4s

Unknown:

Acceleration of the boat = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

        A = \frac{v - u}{t}

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

    Insert the parameters and solve;

  A = \frac{8-8}{4}   = 0m/s²

6 0
2 years ago
Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
Charra [1.4K]

Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

8 0
3 years ago
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