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3 years ago

6

Which of the following is NOT true about sweat? Sweat helps cool the body. Sweat glands are located all over the body. Sweat hel

ps regulate body temperature. Sweat is used to sense pain or pressure changes.

2 answers:

erma4kov [3.2K]3 years ago

8 0

The answer is c!!!!!!!!!!!!!!!

Klio2033 [76]3 years ago

5 0

Sweat is used to sense pain or pressure change.
I eliminated all the ones i wouldnt choose

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In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

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

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

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3 years ago

If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?

Pepsi [2]

Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

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3 years ago

A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t

andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be ( R+ h )

orbital velocity V₀ = $\sqrt{\frac{GM}{( R+h)} }$

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x $\sqrt{\frac{( R+h)}{GM } }$

$\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ = R +h

h = $\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ - R.

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4 years ago

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answer:

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explanation:

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Which statements about electric field lines are correct? Check all that apply.

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