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Nadusha1986 [10]
3 years ago
15

Linda, a biker, is moving along a circular path at a constant speed of 10 km/h. In a neighboring arena Kevin, another biker, is

moving along a circular path at a constant speed of 12 km/h. If the acceleration of the two bikers is the same, what can you infer?
A) The centripetal force on Kevin is higher than that on Linda.
B) The centripetal force on Linda is higher than that on Kevin.
C) Kevin’s circular path has a bigger radius than Linda’s path.
D) Linda’s circular path has a bigger radius than Kevin’s path
Physics
1 answer:
Ulleksa [173]3 years ago
8 0

Answer:

the answer is  C

Explanation:

i took the test just now so i am glad i can help someone in the future❣❣❣❣

<u><em>Also here is the explanation</em></u>

Note that the magnitude of acceleration (a) is given by a =  

v2

R

, where v is the speed and R is the radius. Because acceleration is the same for both bikers, the ratio 

v2

R

, also has to be the same for both. Given that Kevin’s speed is higher, the radius of his path must also be higher to maintain the same ratio. Therefore, Kevin’s circular path has a bigger radius than Linda’s path.

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Answer:

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A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
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Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

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Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
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v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

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\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

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