Answer:Proton (charge of +e, in the nucleus), Neutron (0 charge, in the nucleus), and Electron (charge of –e, outside the nucleus).Nov 13, 2015
Explanation: YA
Answer:
a) Total mass form, density and axis of rotation location are True
b) I = m r²
Explanation:
a) The moment of inertia is the inertia of the rotational movement is defined as
I = ∫ r² dm
Where r is the distance from the pivot point and m the difference in body mass
In general, mass is expressed through density
ρ = m / V
dm = ρ dV
From these two equations we can see that the moment of inertia depends on mass, density and distance
Let's examine the statements, the moment of inertia depends on
- Linear speed False
- Acceleration angular False
- Total mass form True
- density True
- axis of rotation location True
b) we calculate the moment of inertia of a particle
For a particle the mass is at a point whereby the integral is immediate, where the moment of inertia is
I = m r²
v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;
and solve for "v".
______________________________________________________
Explanation:
_____________________________________________________
The formula is: KE = (½) * (m) * (v²) ;
_____________________________________
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since: a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
___________________________________________________
→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
→ We divide each side of the equation by: "[(½)* (m)]" ;
___________________________________________________
→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
to get:
______________________________________________________
→ v² = (KE) / [(½)* (m)]
→ v² = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
→ √ (v²) = √ { 2*(KE) ] / m }
________________________________________________________
→ v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];
and solve for "v".
______________________________________________________
(C) Air Resistance
<u>Explanation:</u>
When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.
Answer:
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Explanation:
First, we must calculate the resultant force (
), in newtons, by vectorial sum:
![\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20%5B%28-200%5C%2CN%29%5Ccdot%20%5Ccos%2060%5E%7B%5Ccirc%7D%2B%28400%5C%2CN%29%5Ccdot%20%5Ccos%2045%5E%7B%5Ccirc%7D%2B300%5C%2CN%5D%5C%2C%5Chat%7Bi%7D%20%2B%20%5B%28200%5C%2CN%29%5Ccdot%20%5Csin%2060%5E%7B%5Ccirc%7D%20%2B%20%28400%5C%2CN%29%5Ccdot%20%5Csin%2045%5E%7B%5Ccirc%7D-100%5C%2CN%5D%5C%2C%5Chat%7Bj%7D)
(1)
Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:
Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:
Where 
is the direction of the resultant force, in sexagesimal degrees.
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
