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svp [43]
3 years ago
15

A student jogs for a mile when their heartbeat starts racing and the student feels too fatigued to keep jogging. The student sto

pped to rest in response to the body's —
A. external stimulus of fatigue.


B. external stimulus of traveling a mile.


C. internal stimulus of sweating.


D. internal stimulus of a increased heart rate.
Physics
2 answers:
Scrat [10]3 years ago
8 0

Answer:

D

Explanation

the asnwer is d

Mama L [17]3 years ago
3 0
D.internal simultaneous of a increased heart rate.
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A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th
tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field  and current given as

B=\dfrac{2\mu _oI}{\pi}D

Where

D=\dfrac{\sqrt{l^2+b^2}}{lb}

By putting the values

D=\dfrac{\sqrt{l^2+b^2}}{lb}

D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}

D=26.03 m⁻¹

B=\dfrac{2\mu _oI}{\pi}D

3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03

I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}

I=1.48 A

8 0
3 years ago
If the frequency of a FM wave is 8.85 × 107 hertz, what is the period of the FM wave?
patriot [66]

Answer:

1.1299 x 10^-8   second    

Explanation:

Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8  sec

8 0
2 years ago
You push a box with a force of 80 n. if the net force on the box is 50 n, what is the force on the box due to sliding friction?
Elodia [21]

The force of the sliding friction is 30 N.

3 0
3 years ago
Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).
JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

6 0
3 years ago
What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?
notka56 [123]
The circumference of the Earth at the equator is listed as 24,901 miles.
So his speed is

                     24,901 miles per day.

Convert it to units that we have a better feel for:

                   (24,901 mi/da) x (1 da / 24 hrs)

             =    (24,901 / 24)  (miles/hour)

             =  about  1,038 miles per hour.

You'll find a huge number of people on the internet these days,
telling you that you could not be moving at that speed and not
feel it, so therefore the Earth is not spinning, and it's not a globe.

I have a lot of feelings and comments about those people, their
lines of reasoning, and their levels of education and intelligence,
so don't get me started.

I just want to guarantee you that everything you're learning about
the Earth and the solar system in school is well founded, and it's
all based on the life's work of some of the smartest people of the
past 300 years of human history.  Everything you're taught about
the Earth has good reasons behind it, whereas those other people
have nothing.

A person on Earth's equator is moving from west to east at roughly
1,038 miles per hour, relative to any point on the Earth's rotation axis.
4 0
3 years ago
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