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AysviL [449]
1 year ago
11

What is the benefit of cooling down after exercise?

Physics
1 answer:
d1i1m1o1n [39]1 year 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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Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl
BARSIC [14]
<span>b. The coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.

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6 0
3 years ago
Which artist of the northern European Renaissance, shown in this self-portrait, was also a block printer and engraver?
kicyunya [14]
It's very hard to see the self-portrait, so I can't identify him.
6 0
3 years ago
Read 2 more answers
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the
Yuri [45]

Answer:

  • <u>The energy change would be 46kJ</u>
  • <u>The energy would be absorbed</u>

Explanation:

The <em>energy change </em>during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the<em> bonds of the products </em>less the chemical energy stored in the <em>bonds of the reactants</em>.

Hence:

  • <em>Energy change</em> = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction <em>absorbed energy</em> and it is endothermic.

4 0
3 years ago
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