Answer:
Too old(Ex. if real time is 1000 then they estimate >1000)
Explanation:
This is because with time our planet may have a definite function which describes temperature.(Because of all the factors and global warming except nuclear bomb testing)
Now nuclear test on planet have significant effect on temperature rise.
Also 14°C rise in temperature is good one because of this.
If future archaeologists only consider that uniform function as above mentioned then they estimate more time then the real one.
Thus too old is right answer.
Lol i think someone would be fired from their job if they threw food
The work done occurs only in the direction the block was moved - horizontally. Work is given by:
W = F(h) * d
Where F(h) is the force applied in that direction (horizontal) and d is the distance in that direction. In this case, F(h) is the horizontal component of the applied force, F(app). However, the question doesn't give us F(app), so we need to find it some other way.
Since the block is moving at a constant speed, we know the horizontal forces must be balanced so that the net force is 0. This means that F(h) must be exactly balanced by the friction force, f. We can express F(h) as a function of F(app):
F(h) = F(app)cos(23)
Friction is a little trickier - since the block is being PUSHED into the ground a bit by the vertical component of the applied force, F(v), the normal force, N, is actually a bit more than mg:
N = mg + F(v) = mg + F(app)sin(23)
Now we can get down to business and solve for F(app) - as mentioned above:
F(h) = f
F(h) = uN
F(h) = u * (mg + F(v))
F(app)cos(23) = 0.20 * (33 * 9.8 + F(app)sin(23))
F(app) = 76.8
Now that we have F(app), we can find the exact value of F(h):
F(h) = F(app)cos(23)
F(h) = 76.8cos(23)
F(h) = 70.7
And now that we have F(h), we can find W:
W = F(h) * d
W = 70.7 * 6.1
W = 431.3
Therefore, the work done by the worker's force is 431.3 J. This also represents the increase in thermal energy of the block-floor system.
Answer:
Explanation:
The period of the comet is the time it takes to do a complete orbit:
T=1951-(-563)=2514 years
writen in seconds:
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](https://tex.z-dn.net/?f=T%5E2%3D%5Cfrac%7B4%5Cpi%5E2%7D%7BGm_%7Bsun%7D%7Da%5E3%5C%5C%20a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGm_%7Bsun%7DT%5E2%7D%7B4%5Cpi%5E2%7D%20%7D%20%5C%5Ca%3D1.50%2A10%5E%7B6%7Dm)
Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:
