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n200080 [17]
3 years ago
11

Why do stretching exercises increase flexibility more than cardio exercises?

Physics
1 answer:
hichkok12 [17]3 years ago
5 0
Stretching exercises increase flexibility more than cardio exercises because stretching exercises stretches your muscles in around your body while cardio exercises only work out the muscles of your cardiovascular system. <span>It will help you in endurance but not so much in flexibility.</span>
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If your heart is not strong enough or efficient enough, it is difficult to (3 points)
Lyrx [107]

it's difficult to breathe.

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3 years ago
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A 30 kg male emperor penguin under a clear sky in the Antarctic winter loses very little heat to the environment by convection;
yarga [219]

Answer:

Rate of energy loss by radiation is 28.31 Watt

Explanation:

Given that;

m = 30 kg

power p = 12 W

emissivity e = 0.97

Surface Area A = 0.56 m²

outside of the penguin's body  T = −22°C

surroundings Temperature Ts = -38°C

the rate of energy loss by radiation = ?

Now, using Stefan-Boltzmann law;

P = σeA [ T⁴ - Ts⁴ ]

Stefan's constant σ = 5.67 × 10⁻⁸

so we substitute

P = 5.67 × 10⁻⁸ × 0.97 ×  0.56  [ (-22 + 273 k)⁴ - (-38 + 273 k )⁴]

= 3.079944 × 10⁻⁸ [ 919325376]

=  28.31 Watt

the rate of energy loss by radiation is 28.31 Watt

8 0
3 years ago
How is 0.00069 written in scienitfic notation?
victus00 [196]
You need to move the decimal point between the six and nine. 6.9 X 10^-4
8 0
3 years ago
Which of the following are advantages of the SI system?
saul85 [17]

Answer:

A is the answer

Explanation:

That's why it ranges from big to small.

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3 years ago
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An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacit
Roman55 [17]

Answer:

a) Energy stored in the capacitor, E = 1.0125 *10^{-3} J

b) Q = 45 µC

c) C' = 1.5 μF

d)  E = 6.75 *10^{-4} J

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J

4 0
3 years ago
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