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3 years ago

What muscle are controlled by the autonomic muscle system and are involuntary

Physics

2 answers:

FromTheMoon [43]3 years ago

4 0

The answer is smooth muscle, Hope this helped! Have a good day!

evablogger [386]3 years ago

3 0

Smooth muscle is the answer.

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2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.

bazaltina [42]

Mass of Jupiter=1.9×10

㎏=M

Mass of Sun=1.99×10

㎏=M

Mean distance of Jupiter from Sun=7.8×10

m=r

G=6.67×10

−11

N㎡㎏

−2

Gravitational Force, F=

(7.8×10

)

6.67×10

−11

×1.9×10

×1.99×10

F=4.16×10

4 0

2 years ago

Can I have some help please :)

Anastasy [175]

Answer:

didya know this is chrating. everyone has a breain and so do u .

Explanation:

use and u will pass

5 0

2 years ago

2. A person standing at

elena-s [515]

God is good man what can you say but its 18.66x30301=362728

7 0

2 years ago

How much conventional current must you run in a solenoid with radius = 0.05 m and length = 0.39 m to produce a magnetic field in

Ugo [173]

Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

$B=\mu _{0}ni$

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

$2\times 10^{-5}=4 \times 3.14\times 10^{-7}\times 512.8\times i$

i = 0.031 A

3 0

2 years ago

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

2 years ago