Answer:
67.9 kg*m/s
Explanation:
Pi = 38 kgm/s
F = 88.3N and ∆t = 0.338s
Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454
=) Pf = 67.8454 kgm/s = 67.85kg*m/s
Your answer is 67.9kg*m/s with three significant figures
hope this helps your troubles!
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
Answer:
Explanation:
1. We can find the temperature of each star using the Wien's Law. This law is given by:
![\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}](https://tex.z-dn.net/?f=%5Clambda_%7Bmax%7D%3D%5Cfrac%7Bb%7D%7BT%7D%3D%5Cfrac%7B2.9x10%5E%7B-3%7D%5BmK%5D%7D%7BT%5BK%5D%7D)
(1)
So, the temperature of the first and the second star will be:
Now the relation between the absolute luminosity and apparent brightness is given:
(2)
Where:
- L is the absolute luminosity
- l is the apparent brightness
- r is the distance from us in light years
Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂
If we use the equation (2) we have:
So the relative distance between both stars will be:
(3)
The Boltzmann Law says, 
(4)
- σ is the Boltzmann constant
- A is the area
- T is the temperature
- L is the absolute luminosity
Let's put (4) in (3) for each star.
As we know both stars have the same size we can canceled out the areas.
I hope it helps!
<span>A decrease in the overall volume of gases namely hydrogen would prevent nuclear fusion in a nebula.</span>
