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kirill115 [55]
2 years ago
14

Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R

oss takes two months off from training what will likely occur?
Physics
1 answer:
damaskus [11]2 years ago
4 0

If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

  • In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.
  • Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.
  • A decrease in strength and muscle mass, with beginners experiencing a smaller decline in strength than experienced lifters.
  • Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.

learn more about fitness here: brainly.com/question/13490156

#SPJ10

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Answer:

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Explanation:

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3 years ago
The black lines that wrap themselves around a basketball represent an example of what type of geometry?
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In Euclidean geometry parallel lines never intersect. But in non-Euclidean geometry parallel lines can either curve away from each other, or curve towards each other. Example : the black lines that wrap themselves around the basketball.
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How many atoms are in a sample of 2.39 moles of neon (Ne) atoms?
Reil [10]
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3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
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3 years ago
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B. Atoms are solid balls
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