Answer:
![\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B24%7D%7B1.14375%7D%3D20.983%5Cfrac%7Brad%7D%7Bs%7D)
Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:
![H_o =r x mv=rxL](https://tex.z-dn.net/?f=H_o%20%3Dr%20x%20mv%3DrxL)
Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:
![\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1](https://tex.z-dn.net/?f=%5Cint_%7Bt_1%7D%5E%7Bt_2%7DM_O%20dt%20%3D%20%5Cint_%7Bt_1%7D%5E%7Bt_2%7DH_O%20dt%3DH_0t2%20-H_0t1)
Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is
".
If we analyze the staritning point we see that the initial velocity can be founded like this:
![v_o =\omega r_{OIC}=\omega (0.15m)](https://tex.z-dn.net/?f=v_o%20%3D%5Comega%20r_%7BOIC%7D%3D%5Comega%20%280.15m%29)
And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:
![H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af](https://tex.z-dn.net/?f=H_Ai%20%2B%5Csum%20%5Cint_%7Bt_i%7D%5E%7Bt_f%7D%20M_A%20dt%20%3DH_Af)
![0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have
![1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega](https://tex.z-dn.net/?f=1.5%284%5E2%29-1.5%280%5E2%29%20%3D%200.46875%5Comega%20%2B0.675%5Comega%3D1.14375%5Comega)
And if we solve for
we got:
![\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%5Cfrac%7B24%7D%7B1.14375%7D%3D20.983%5Cfrac%7Brad%7D%7Bs%7D)
Answer:
la escuela,en casa y listo...............
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3
Explanation:
The step by step and appropriate substitution is as shown in the attachment.
From number of moles = Concentration x volume
number of moles = number of particles/ Avogadro's number
Volume = mass/density, the appropriate derivation to get the number of moles of atoms
I don’t know what you mean by that