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ss7ja [257]
3 years ago
11

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

tory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff if air resistance is negligibly small?
Physics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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3 years ago
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I think the correct answer is is D.
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3 years ago
A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax
levacccp [35]

Answer:

\alpha = 6.431\,\frac{rad}{s^{2}}

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The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

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Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

T = \frac{1}{2}\cdot M \cdot R \cdot \alpha

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\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}

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7 0
3 years ago
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Oduvanchick [21]

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THE GASEOUS STATE
Pressure  atm
Volume  liters
n  moles
R  L atm mol^-1 K^-1
Temperature  Kelvin


pv = rt

divide both sides by v
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answer: p = rt/v




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PV = NRT
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mass/V = P (MW)/RT = density



Molar Mass:
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For Ideal Gasses:


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= n_ A/n_ total = X_ A





Therefore, P_ A = X_ A P_ total.



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V volume


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Hope that helps!!!!!! Have a great day : )

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Answer:

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Explanation:

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