The force of gravity between the astronauts is ![1.67\cdot 10^{-7}N](https://tex.z-dn.net/?f=1.67%5Ccdot%2010%5E%7B-7%7DN)
Explanation:
The magnitude of the gravitational force between two objects is given by:
where
:
is the gravitational constant
are the masses of the two objects
r is the separation between them
In this problem, we have two astronauts, whose masses are:
![m_1 = 100 kg\\m_2 = 100 kg](https://tex.z-dn.net/?f=m_1%20%3D%20100%20kg%5C%5Cm_2%20%3D%20100%20kg)
While the separation between the astronauts is
r = 2 m
Substituting into the equation, we can find the gravitational force between the two astronauts:
![F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%286.67%5Ccdot%2010%5E%7B-11%7D%29%28100%29%28100%29%7D%7B2%5E2%7D%3D1.67%5Ccdot%2010%5E%7B-7%7DN)
Learn more about gravitational force:
brainly.com/question/1724648
brainly.com/question/12785992
#LearnwithBrainly
Answer:
![a=16\ m/s^2](https://tex.z-dn.net/?f=a%3D16%5C%20m%2Fs%5E2)
Explanation:
<u>Motion With Constant Acceleration
</u>
It's a type of motion in which the velocity of an object changes uniformly in time.
The formula to calculate the change of velocities is:
![v_f=v_o+at](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat)
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:
![\displaystyle a=\frac{v_f-v_o}{t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7Bv_f-v_o%7D%7Bt%7D)
![\displaystyle a=\frac{70-6}{4}=\frac{64}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B70-6%7D%7B4%7D%3D%5Cfrac%7B64%7D%7B4%7D)
![\boxed{a=16\ m/s^2}](https://tex.z-dn.net/?f=%5Cboxed%7Ba%3D16%5C%20m%2Fs%5E2%7D)
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other.
, the amount of charge stored in this capacitor, will stay the same.
The formula
relates the electric potential across a capacitor to:
, the charge stored in the capacitor, and
, the capacitance of this capacitor.
While
stays the same, moving the two plates apart could affect the potential
by changing the capacitance
of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of
. Neither will that change the area of the two plates.
However, as
(the distance between the two plates) increases, the value of
will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula
can be rewritten as:
.
The value of
(charge stored in this capacitor) stays the same. As the value of
becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Valence electrons are required for certain bonding processes. Without them you cannot bond with certain elements. For example, carbon bonds really well with carbon because it has the same amount of valence electrons. However carbon would not bond well with uranium due to the massive differences in valence electrons. Hope this helps!
Answer:
It would be PE=16kg * 9.8 m/s^2 * 1m = 160 J
Explanation:
The person who asked this question ended up answering his own question so I'm here to let you know all that the answer was founded by the person whos posted the question himself full credit goes to him :)