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2 years ago

Is using a heater to keep warm when it is cold conduction convection or radiation

Physics

1 answer:

Dmitry_Shevchenko [17]2 years ago

8 0

Answer: convection

Explanation:

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A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the e

Tom [10]

Answer:

a) $F=16741.9N$

b) $\frac{F}{W}=0.795$

Explanation:

The gravitational force on the satellite is calculated with Newton's Gravitation Law:

$F=\frac{GMm}{r^2}$

Where $M=5.97\times10^{24}kg$ is Earth's mass, $m=2150kg$ is the satellite mass, $r=R+h$ is the distance between their centers, where $h=780000m$ is the height of the satellite (from Earth's surface) and $R=6371000m$ is Earth's radius, and $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

a) With these values we then have:

$F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N$

b) And the fraction this force is of the satellite’s weight <em>W=mg</em> is:

$\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795$

5 0

4 years ago

Infuse 1000 mL of NS. The infusion set delivers 15 gtt/mL. Give 300 mL bolus over 20 minutes and then infuse at the hourly rate

Bas_tet [7]

Answer:

rate of infusion is 900 mL/hr

Explanation:

given data

Infuse I1 = 1000 mL

delivers = 15 gtt/mL

Infuse I2 = 300 mL

time t= 20 min

rate = 60 mL/hr

to find out

rate of infusion

solution

we know here we give 300 mL infuse in 20 min

so here for 20 min

rate of infusion is express as

rate of infusion = I2 / t

rate of infusion = 300 / 20

rate of infusion = 15 mL / min

rate of infusion = 15 × 60 = 900

so rate of infusion is 900 mL/hr

4 0

3 years ago

What is used to make a cast of a tool mark ?

KatRina [158]

silicone rubber is used to make a cast of a tool mark.

8 0

3 years ago

Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob

Mars2501 [29]

Answer:

$a=2.77*10^{13}m$

$R_a=5.49*10^{13}m$

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

$2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s$

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

$T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m$

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

$R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m$

8 0

3 years ago

Ahannwnsnwn dnsjsn and

bazaltina [42]

I dont know what you are trying to say but okay

4 0

3 years ago