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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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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 5.00 s, it rotates 13.9 rad. Du
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Answer:

(a) Angular acceleration is 1.112 rad/s².

(b) Average angular velocity is 2.78 rad/s .

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The equation of motion in Rotational kinematics is:

θ = θ₀ + 0.5αt²

Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.

(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:

13.9 = 0 + 0.5α(5)²

α = 1.112 rad/s²

(b) The equation of average angular velocity is:

ω = Δθ/Δt

ω = \frac{13.9}{5}

ω = 2.78 rad/s

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3 years ago
A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second
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The power that the light is able to utilize out of the supply is only 0.089 of the given.
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3 years ago
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A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

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c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

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The Green work gymno means Naked
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