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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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Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati

jek_recluse [69]

Answer:

$a=4\ m/s^2$

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2$

So, the magnitude of centripetal acceleration is $4\ m/s^2$ .

5 0

2 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to -207.5kW

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

Which of the following does NOT represent Newton’s second law? Question 20 options: a = m/Fnet m = Fnet/a Fnet = ma a = Fnet/m

Natali [406]

Answer:

a=m/f is not an equation under newton's second law

Explanation:

newton's second law of motion is represented using: f=ma

where a=v-u/t

therefore it becomes,f=m(v-u)/t

from f=ma,

a will become f/m,

m will become f/a

8 0

3 years ago

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

The number of fringe is z = 3 fringes

The ratio is $I = 0.2545I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$