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3 years ago

Why are slow-twitch muscles more beneficial than fast-twitch muscles for cardiorespiratory fitness?

Physics

2 answers:

nata0808 [166]3 years ago

6 0

The answer is D. Slow-twitch muscles are able to use oxygen more efficiently than fast-twitch muscles.

Hope this helps

dlinn [17]3 years ago

4 0

Answer:D

Explanation: with fitness and running you normally run for longer distances the slow twitch mussels help with this by using less oxygen in a longer amount of time

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Which of the materials would be able to scratch Quartz

Lelechka [254]

<span>anything harder than mohs scale 7 so eg Topaz, Corundum and diamond representing mohs scale 8 9 and 10 respectively.</span>

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3 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

For more information on this visit

brainly.com/question/21811998

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3 years ago

The two loudspeakers in the drawing are producing identical sound waves. The waves spread out and overlap at the point P. What i

Damm [24]

Answer:

5/2π

Explanation:

According to quizlet the answer is 5/2π

5 0

4 years ago

Suppose the ends of a 27-m-long steel beam are rigidly clamped at 0°C to prevent expansion. The rail has a cross-sectional area

frozen [14]

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

$\Delta T = 39$ Degree celcius

change in length of steel beam is given as

$\Delta L = L_O \alpha \Delta T$

$= 20*1.1*10^{-5}*39$

$=8.58*10^{-3}$ m

Young's modulus is

$Y = \frac{FL}{A\Delta L}$

$F = \frac{ YA\Delta L}{L}$

$= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}$

F = 1.58*10^{11} N

5 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find i

serg [7]

Answer:

$\frac{1}{12}ML^2$

Explanation:

The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:

$I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2$

$I = \frac{1}{3}M\frac{L^2}{4}$

$I = \frac{1}{12}ML^2$

5 0

3 years ago