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Mandarinka [93]
3 years ago
5

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. the acceleration

of gravity is 9.8 m/s 2 . if her speed is 430 m/s at this time, find the radius of the loop. answer in units of km.
Physics
1 answer:
Yanka [14]3 years ago
6 0
The acceleration due to gravity serves as the centripetal acceleration of the objects that orbits the Earth. The centripetal acceleration due to gravity is calculated through the equation,

    a = v²/r

where v is the speed and r is the radius. Substituting the known values to the equation,

   9.8 m/s² = (420 m/s)² / r

The value of r from the equation is 18000 m or equal to 18 km.

<em>Answer: 18 km</em>
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Explanation:

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Case 1.

R = 60 m v = 16 m/s

F=\dfrac{(16)^2m}{60}=4.26m\ N

Case 2.

R = 15 m v = 8 m/s

F=\dfrac{(8)^2m}{15}=4.26m\ N

Case 3.

R = 30 m v = 4 m/s

F=\dfrac{(4)^2m}{30}=0.54m\ N

Case 4.

R = 45 m v = 4 m/s

F=\dfrac{(4)^2m}{45}=0.36m\ N

Case 5.

R = 30 m v = 16 m/s

F=\dfrac{(16)^2m}{30}=8.54m\ N

Case 6.

R = 15 m v =12 m/s

F=\dfrac{(12)^2m}{15}=9.6m\ N

Ranking from largest to smallest is given by :

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5 0
3 years ago
What is the cell emf for the concentrations given? Express your answer using two significant figures.
alisha [4.7K]

Complete Question

A  voltaic cell is  constructed with two Zn^{2+}- Zn electrodes. The  two cell compartment have  [Zn^{2+}] =  1.6 \ M and  [Zn^{2+}] =  2.00*10^{-2} \  M respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is   E =  0.06 V

Explanation:

Generally from the question we are told that

   The  concentration of [Zn^{2+}] at the cathode is  [Zn^{2+}]_a =  1.6 \ M

    The  concentration of [Zn^{2+}] at the anode is [Zn^{2+}]_c =  2.00*10^{-2} \  M

Generally the the cell emf for the concentration is mathematically represented as

     E =  E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}

Generally the E^ois the standard emf of a cell, the value is  0 V

So

      E =  0  -  \frac{0.0591}{2}  * log[\frac{ 2.00*10^{-2}}{1.6} ]

=>      E =  0.06 V

4 0
3 years ago
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