Answer:
See explanation
Explanation:
We have a mass 
revolving around an axis with an angular speed 
, the distance from the axis is 
. We are given:
![\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]](https://tex.z-dn.net/?f=%5Comega%20%3D%2010%20%5Brad%2Fs%5D%5C%5Cr%3D0.5%20%5Bm%5D%5C%5Cm%3D13%5BKg%5D)
and also the formula which states that the kinetic rotational energy of a body is:
.
Now we use the kinetic energy formula
where 
is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:
After replacing in the previous equation we get:
now we have the following:
therefore:
then the moment of inertia will be:
![I = 13*(0.5)^2=3.25 [Kg*m^2]](https://tex.z-dn.net/?f=I%20%3D%2013%2A%280.5%29%5E2%3D3.25%20%5BKg%2Am%5E2%5D)
Answer:
can you show a graph but if not i believe the answer is x=6m
Explanation:
Crossing over, or recombination, is the exchange of chromosome segments between nonsister chromatids in meiosis. Crossing over creates new combinations of genes in the gametes that are not found in either parent, contributing to genetic diversity.
Answer:
10
Explanation:
The decibel or decibel, is a unit that relates two values of sound pressure, and electrical power. the base unit is the bel, but given the amplitude but for practicality, a submultiple, the decibel, is used. It is a scalar expression .
Zero belios is the reference value used as a base, hence a bel is equivalent to 10 decibels and means that there is a power increase of 10 times over the reference value.
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />![F_{y}=2[N]](https://tex.z-dn.net/?f=F_%7By%7D%3D2%5BN%5D)
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />![F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D-1%2Asin%2860%29%5C%5CF_%7Bx%7D%3D-0.866%5BN%5D%5C%5CF_%7By%7D%3D1%2Acos%2860%29%5C%5CF_%7By%7D%3D0.5%20%5BN%5D)
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)
