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3 years ago

Match the major muscle group with its functional role.

Physics

1 answer:

Naily [24]3 years ago

7 0

Answer:

Abdominal

Sitting up, postural alignment

Biceps

Lifting, pulling

Deltoids

Overhead lifting

Erector Spinae

Postural alignment

Gastronemius & Soleus

Push off for walking, standing on tiptoes

Gluteus

Climbing stairs, walking, standing up

Hamstrings

Walking

Latissimus Dorsi & Rhomboids

Postural alignment, pulling open a door

Obliques

Rotation and side flexion of body

Pectoralis

Push up, pull up, bench press

Quadriceps

Climbing stairs, walking, standing up

Trapezius

Moves head sideways

Triceps

Pushing

God bless you. Because my soul almost left my body when i had to do this.

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(a) Calculate the self-inductance (in mH) of a 55.0 cm long, 10.0 cm diameter solenoid having 1000 loops.

DedPeter [7]

Explanation:

(a) We have,

Length of solenoid, l = 55 cm = 0.55 m

Diameter of the solenoid, d = 10 cm

Radius, r = 5 cm = 0.05 m

Number of loops in the solenoid is 1000.

(a) The self inductance in the solenoid is given by :

$L=\dfrac{\mu_o N^2A}{l}$

A is area

$L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH$

(b) The energy stored in the inductor is given by :

$E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J$

Hence, this is the required solution.

4 0

3 years ago

Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0

3 years ago

A 1800-kg Jeep travels along a straight 500-m portion of highway (from A to B) at a constant speed of 10 m/s. At B, the Jeep enc

vladimir1956 [14]

Answer:

3600N

Explanation:

$F=m\frac{v^2}{r}$

4 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

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6 0

3 years ago

In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?

Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

negative charges
positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

#SPJ1

5 0

1 year ago