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Nadya [2.5K]
3 years ago
15

What is weight if m is mass and g is gravitational force.

Physics
1 answer:
Nataly_w [17]3 years ago
7 0
Weight is a force, and is defined as mass times acceleration. So (g is acceleration and m is mass):
F_g=mg

That's your answer.
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Help please!!!
exis [7]

Answer:

5 years worth of work (aka all of the homework i currently have)

3 0
3 years ago
Which is true regarding the penetrating power of radiation? (2 points) Select one:
Monica [59]

Answer:d

Explanation:

Alpha particles are heaviest among alpha, beta and gamma so they have least amount of Penetration compared to both.

Gamma Particles are lightest among three so they can Penetrate most .

The order of Penetration is given by

Alpha< Beta < Gamma              

3 0
3 years ago
Bill leaves his 60 W desk lamp on every day, including weekends, for eight hours. After one month (30 days), how much total ener
maxonik [38]

' W ' is the symbol for 'Watt' ... the unit of power equal to 1 joule/second.

That's all the physics we need to know to answer this question.
The rest is just arithmetic.

(60 joules/sec) · (30 days) · (8 hours/day) · (3600 sec/hour)

= (60 · 30 · 8 · 3600) (joule · day · hour · sec) / (sec · day · hour)

= 51,840,000 joules
__________________________________

Wait a minute !  Hold up !  Hee haw !  Whoa ! 
Excuse me.  That will never do.
I see they want the answer in units of kilowatt-hours (kWh).
In that case, it's

(60 watts) · (30 days) · (8 hours/day) · (1 kW/1,000 watts)

= (60 · 30 · 8 · 1 / 1,000) (watt · day · hour · kW / day · watt)

= 14.4 kW·hour

Rounded to the nearest whole number:

14 kWh

7 0
3 years ago
Urgente!!!!! <br> Necesito ayuda con esto!!!!
ahrayia [7]
media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4
5 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
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