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Lera25 [3.4K]
3 years ago
13

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

es on at 6.5 m/s. What is her acceleration on the rough ice?
Physics
1 answer:
malfutka [58]3 years ago
8 0

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

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Answer:

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B >>>>>>>> 8 >>>>>>>>> 12

From the table above we can see that both atoms have the same proton number.

Therefore, they are the same element because they have the same proton number which means that they have the same atomic number. The element in this case is existing as an isotope in that the atoms have the same proton number but different neutron number.

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