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3 years ago

13

The ability of a muscle group to exert sub-maximal forces against a resistance over numerous contractions is called A. muscle en

durance. B. muscle overload. C. muscle progression. D. muscle strength.

1 answer:

Delicious77 [7]3 years ago

8 0

The answer is:
A) Muscle Endurance

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At a boundary, plates move away from each other as magma pushes them from below. When this change happens beneath the ocean floo

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The plates under the ocean squeeze really hard and they can start an earthquake.

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4 years ago

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You are observing a star that is 250 million light years away. From the starlight you are observing now, the appears to be 30 mi

Romashka [77]

What ever games your playing it wont work you cant defeat me hahhahha oh I know :O but he can AHHHH :D D: NOOOO IMPOSSIBLE.

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3 years ago

We have all complained that there aren't enough hours in a day. In an attempt to fix that, suppose all the people in the world l

Natali5045456 [20]

Answer:

the duration of a day increases 4.96x10^-11 s

Explanation:

According the exercise:

R=radius of Earth=6.37x10^6 m

mE=mass of Earth=5.97x10^24 kg

m=average mass of people=55 kg

n=number of population=7x10^9

M=total mass=n*m=7x10^9*55=3.85x10^11 kg

v=speed=2.5 m/s

The moment of inertia of population is:

I=(2/3)*M*R^2=(2/3)*3.85x10^11*(6.37x10^6)=1.04x10^25 kg*m^2

The time taken per revolution is:

T=2πR/v=(2π*6.37x10^6)/2.5=1.6x10^7 rev/s

The angular speed is:

w=2π/T=2π/1.6x10^7=3.9x10^-7 rad/s

The angular momentum of population is equal to:

L1=I*w=1.04x10^25*3.9x10^-7=4.08x10^18 kg*m^2/s

The angular momentum of Earth is equal to:

L2=I*w=((2/5)*me*R^2)*(2π/24)=((2/5)*5.97x10^24*(6.37x10^6)^2)*(2π/(24*60*60))=7.1x10^33 kg*m^2/s

The change in length of the day is equal to:

T´=T*(L1/L2)=(24*60*60)*(4.08x10^18/7.1x10^33)=4.96x10^-11 s

8 0

3 years ago

What is the intensity level of a sound with an intensity of 0.000127 W/m2?

irina1246 [14]

DB = 10 log (I/Io)
Where, Io is the reference intensity which is the minimum intensity that human can hear.
That is; Io = 10^-12
Therefore;
dB = 10 log (0.000127/10^-12) =80.79
= 81 dB
Therefore; the correct answer is 81 dB

6 0

4 years ago

An alternator consists of a coil of area A with N turns that rotates in a uniform field B around a diameter perpendicular to the

matrenka [14]

Answer:

(a): $\rm 2\pi f\ NBA\ \sin(2\pi ft).$ .

(b): 20.94 Volts.

Explanation:

<u>Given:</u>

Area of the coil = A.
Number of turns of the coil = N.
Magnetic field in which the coil is placed = B.
The frequency with which the coil is rotating = f.

<h2>(a):</h2>

The magnetic flux linked with a coil is defined as

$\phi = N\vec B \cdot \vec A=\rm NBA\cos\theta$

where,

$\vec A$ is the area vector of the coil, directed along the normal to the plane of the coil.

$\theta$ = angle between the magnetic field and the area vector of the coil.

Assuming that the magnetic field is along the normal to the plane of the coil, initially.

Therefore, at any later time t, the angle which the magnetic field makes with the normal to the plane of the coil is is given by

$\rm \theta = 2\pi ft.$

Therefore, the magnetic flux linked with the coil at any time t is given by

$\rm \phi = NBA\cos(2\pi ft).$

According to the Faraday's law of electromagnetic induction, the emf induced in the coil is given by

$\rm e=-\dfrac{d\phi}{dt}\\=-\dfrac{d(NBA\cos(2\pi ft))}{dt}\\=-NBA\dfrac{d(\cos(2\pi ft))}{dt}\\=-NBA(-2\pi f\sin(2\pi ft))\\=2\pi f\ NBA\ \sin(2\pi ft).$

<h2>(b):</h2>

The amplitude of the alternating voltage is the maximum value of the induced emf in the coil, the induced emf in the coil is maximum when $\rm \sin(2\pi ft)=1$ .

Therefore, the amplitude of the alternating voltage is given by

$\rm e_o=(2\pi f)NBA\ 1\\=2\pi ft \ NBA.$

Given values are:

N = 100 turns.
A = $\rm 10^{-2}\ m^2$ .
B = 0.1 T.
f = 2000 rev/min = $\rm 2000\times \dfrac1{60}\ rev/s = 33.33\ rev/s.$

Putting all these values,

$\rm e_o=2\pi \times 33.33\times (100)\times (0.1)\times (10^{-2})=20.94\ Volts.$

7 0

3 years ago