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3 years ago

How are slow twitch fibers and fast twitch fibers different?

1 answer:

4 0

Slow-twitch muscles help enable long-endurance feats such as distance running.
fast-twitch muscles fatigue faster but are used in powerful bursts of movements like sprinting.

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Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

Rashid [163]

Answer:

a) 0.72 V

b) 19.2 mA

c) 2.304 Watts

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = $N_{p}$ = 500 turns

input voltage = $V_{p}$ = 120 V

number of secondary turns = $N_{s}$ = 3 turns

output voltage = $V_{s}$ = ?

using the equation for a transformer

$\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }$

substituting values, we have

$\frac{V_{s} }{120 } = \frac{3 }{500} }$

$500V_{p} = 120*3\\500V_{p} = 360$

$V_{p}$ = 360/500 = 0.72 V

b) by law of energy conservation,

$I_{P}V_{p} = I_{s}V_{s}$

where

$I_{p}$ = input current = ?

$I_{s}$ = output voltage = 3.2 A

$V_{s}$ = output voltage = 0.72 V

$V_{p}$ = input voltage = 120 V

substituting values, we have

120 $I_{p}$ = 3.2 x 0.72

120 $I_{p}$ = 2.304

$I_{p}$ = 2.304/120 = 0.0192 A

= 19.2 mA

c) power input = $I_{p} V_{p}$

==> 0.0192 x 120 = 2.304 Watts

7 0

3 years ago

Metallurgy is the study of _____.

vitfil [10]

Forming ionic bonds study

6 0

3 years ago

Read 2 more answers

What is the half life-life of your 100 atoms of Carbon-14?

Naddika [18.5K]

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

3 0

2 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

In order for a person to "see" an object, light waves pass through the ___________.

spayn [35]

I think the correct answer from the choices listed above is the first option. In order for a person to "see" an object, light waves pass through the cornea. The cornea is the transparent layer forming at the front of the eye. Hope this answers the question. Have a nice day.

8 0

3 years ago

Read 2 more answers