Well as we all know Atoms have the same number of electrons and protons the net has 0. When an Atom gains or loses electrons it becomes ion when electrons are donated from ion it becomes Cation.
Cation is your correct answer.
Hope This Helps!
Answer:
b) It is impossible to tell without knowing the masses.
Explanation:
The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by
![\Delta T= \frac{Q}{m C_s}](https://tex.z-dn.net/?f=%5CDelta%20T%3D%20%5Cfrac%7BQ%7D%7Bm%20C_s%7D)
where
Q is the amount of heat
m is the mass of the substance
Cs is the specific heat capacity of the substance
In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.
To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.
The angular velocity can be described as
![\omega_f = \omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_0%20%2B%20%5Calpha%20t)
Where,
Final Angular Velocity
Initial Angular velocity
Angular acceleration
t = time
The relation between the tangential acceleration is given as,
![a = \alpha r](https://tex.z-dn.net/?f=a%20%3D%20%5Calpha%20r)
where,
r = radius.
PART A ) Using our values and replacing at the previous equation we have that
![\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%2894rpm%29%28%5Cfrac%7B2%5Cpi%20rad%7D%7B60s%7D%29%3D%209.8436rad%2Fs)
![\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s](https://tex.z-dn.net/?f=%5Comega_0%20%3D%2063rpm%28%5Cfrac%7B2%5Cpi%20rad%7D%7B60s%7D%29%3D%206.5973rad%2Fs)
![t = 11s](https://tex.z-dn.net/?f=t%20%3D%2011s)
Replacing the previous equation with our values we have,
![\omega_f = \omega_0 + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_0%20%2B%20%5Calpha%20t)
![9.8436 = 6.5973 + \alpha (11)](https://tex.z-dn.net/?f=9.8436%20%3D%206.5973%20%2B%20%5Calpha%20%2811%29)
![\alpha = \frac{9.8436- 6.5973}{11}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B9.8436-%206.5973%7D%7B11%7D)
![\alpha = 0.295rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.295rad%2Fs%5E2)
The tangential velocity then would be,
![a = \alpha r](https://tex.z-dn.net/?f=a%20%3D%20%5Calpha%20r)
![a = (0.295)(0.2)](https://tex.z-dn.net/?f=a%20%3D%20%280.295%29%280.2%29)
![a = 0.059m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.059m%2Fs%5E2)
Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation
![\omega_f^2=\omega_0^2+2\alpha\theta](https://tex.z-dn.net/?f=%5Comega_f%5E2%3D%5Comega_0%5E2%2B2%5Calpha%5Ctheta)
Replacing with our values and re-arrange to find ![\theta,](https://tex.z-dn.net/?f=%5Ctheta%2C)
![\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B%5Comega_f%5E2-%5Comega_0%5E2%7D%7B2%5Calpha%7D)
![\theta = \frac{9.8436^2-6.5973^2}{2*0.295}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B9.8436%5E2-6.5973%5E2%7D%7B2%2A0.295%7D)
![\theta = 90.461rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090.461rad)
That is equal in revolution to
![\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090.461rad%28%5Cfrac%7B1rev%7D%7B2%5Cpi%20rad%7D%29%20%3D%2014.397rev)
The linear displacement of the system is,
![x = \theta*(2\pi*r)](https://tex.z-dn.net/?f=x%20%3D%20%5Ctheta%2A%282%5Cpi%2Ar%29)
![x = 14.397*(2\pi*\frac{0.25}{2})](https://tex.z-dn.net/?f=x%20%3D%2014.397%2A%282%5Cpi%2A%5Cfrac%7B0.25%7D%7B2%7D%29)
![x = 11.3m](https://tex.z-dn.net/?f=x%20%3D%2011.3m)
Answer: Thus, the force is directed 72.5° above the horizontal.
Explanation:
The magnitude of the force, F = 15 N
let this force be detected at angle θ from the horizontal, then horizontal component of force is: F cos θ
and vertical component is F sin θ
Horizontal component is given,
F cos θ = 4.5 N
⇒15 N cos θ = 4.5 N
⇒ cos θ = 0.3
⇒ θ = cos⁻¹ 0.3 = 72.5°