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3 years ago

There are five different types of muscles in exercise physiology.

Physics

1 answer:

9966 [12]3 years ago

3 0

Answer: The answer is B) False

Explanation: There are only three muscle types in the exercise physiology.

Skeletal Muscle

Smooth Muscle

Cardiac Muscle

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Help with questions 1 a (I) and 1 a (I)

Cloud [144]

Vibrations with troughs like a light wave

when you touch something that is stringy and wet/dry and move the object acts like a slinky of some sort

3 0

3 years ago

The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii

tankabanditka [31]

Answer:

$0.2528$

Explanation:

To calculate the period we need the formula:

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

Where $r$ is the radius of the moon, $G$ is the universal constant of gravitation and $M$ is the mass of mars.

The period of Phobos:

$T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}$

The period of Deimos:

$T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}$

The ratio of the period of Phobos and Deimos:

$\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}$

$\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}$

Most terms get canceled and we have:

$\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}$

According to the problem

$r_{p}=9,378km\\r_{D}=23,459km$

so the ratio will be:

$\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275$ ≈ $0.2528$

the ratio of the period of revolution of Phobos to that of Deimos is 0.2528

8 0

3 years ago

A 75kg man goes up a tower 30 m in 120s. How much power did the man exert

Pepsi [2]

Explanation:

If g= 10m/s²

Then 75kg=75×10=750N

Since Work =Force ×Distance

Work=750×30

=22500J

And Power°=Work÷time

=22500÷120

=187.5W

8 0

3 years ago

How long will it take a car to go from a complete stop to 44 km/hr if they are accelerating at 5 km/hr?

Aleonysh [2.5K]

Solution :-

Given that ,

Initial Velocity of car = 44km / hr.
Final Velocity of car = 0km / hr.
Acceleration = -5km/hr².

To Find :

Time taken to stop the car .

So , here use first equⁿ of motion which is ,

$\boxed{\red{\bf\dag v = u + at }}$

where ,

v is final Velocity.
u is Initial velocity.
a is acceleration.
t is time taken.

Now , substituting the respective values ,

$\tt:\implies v=u+at$

$\tt:\implies 0 = 44 + (-5)t$

$\tt:\implies -44 = -5t$

$\tt:\implies t=\dfrac{44}{5}$

$\underline{\boxed{\red{\tt \longmapsto Time\:\:=\:\:8.8hrs.}}}$

$\boxed{\green{\bf\pink{\dag} Hence\:time\:taken\:to\:stop\:the\:car\:is\:8.8hrs.}}$

6 0

3 years ago

Which is less likely to be a reliable source of information, the webpage of a university or the webpage of a scientist who is tr

tatyana61 [14]

webpage of a scientist who is trying to sell a new invention

trying to dell = vested interest.

umiversity should be objective impartial

7 0

3 years ago