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AysviL [449]
2 years ago
11

What is the benefit of cooling down after exercise?

Physics
1 answer:
d1i1m1o1n [39]2 years ago
3 0

Answer:

3 benefits of a restrict afterwards a practice Brings essence rate back off and slows respiring Muscles return back to optimum motionless distance Prevents ancestry from combining in lower limits Cool down stretches are used to weaken courage rate and respiring rates afterwards exercise. It can help cause your soul rate back off to situated rate.

Explanation:

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The higher the value of the coefficient of friction, the _______ the resistance to sliding.
Tom [10]
<span>The higher the value of the coefficient of friction, the more the resistance to sliding.  The answer is the more the resistance to sliding.  The</span> coefficient of friction<span> is a measure of how easily one object moves over another object. It is a ratio of: Force to move the object / weight of the object (or Normal Force)</span>
7 0
3 years ago
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A semi has a net force of 13,000 N and is accelerating at a rate of 6.5 m/s^2. What is the mass of the semi? *
Evgen [1.6K]

Example 1: An 850-kg car is accelerating at a rate of 2.4m/s2 to the right along a ... 2) A nonzero net force ΣF acting on mass M causes an acceleration a in it such that ΣF = Ma. The acceleration has the same direction as the applied net force. ... (b) Knowing that the crate is being pushed to the left by a 53-N force

4 0
3 years ago
How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and
aev [14]

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h           ( g = 10 m/s²)

h= 46.66 m

4 0
3 years ago
W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively
MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c)  W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

 W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

      W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

    W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

      ΔU = ΔK

7 0
3 years ago
Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 3 s. What is your acceleration?
lisov135 [29]

You asked the question twice I answered it on the last one

7 0
3 years ago
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