Answer:
The frequency of the oscillations in terms of fo will be f2=fo/3
E xplanation:
T= 
=1:3
⇒f2=fo\3
Here frequency f is inversely poportional to square root of mass m.
so the value of remainder of frequency f2 and fo is equal to 1:3.
⇒
= 
⇒
= 1:3
⇒f2=
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force
acting against the motion. Since their resultant must be zero, we have:
The frictional force is
where
is the coefficient of kinetic friction
is the weight of the block.
Substituting these values, we find the magnitude of the force F:
The answer is D. wavelength!
Answer:
According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."
Explanation:
Hope this helps! :)
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
