Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire,
=
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, ![n = \frac{\rho}{M}N_A](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B%5Crho%7D%7BM%7DN_A)
where,
=Avogadro's Number
![n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B8.93%7D%7B63.5%7D%286.2%5Ctimes%2010%5E%7B23%7D%29%3D8.719%5Ctimes%2010%5E%7B28%7D%5C%20electrons%2Fm%5E3)
Now,
the drift velocity, ![V_d](https://tex.z-dn.net/?f=V_d)
![V_d=\frac{J}{ne}](https://tex.z-dn.net/?f=V_d%3D%5Cfrac%7BJ%7D%7Bne%7D)
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Answer: B, Companies passed on production and transportation costs to consumers
Explanation:
Answer:
The angle of banked curve that makes the reliance on friction unnecessary is
![\arcsin(v^2/(gR))](https://tex.z-dn.net/?f=%20%5Carcsin%28v%5E2%2F%28gR%29%29)
Explanation:
In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.
The only force in the direction of radius is the sine component of the weight of the car
![w_r = mg\sin(\theta)](https://tex.z-dn.net/?f=w_r%20%3D%20mg%5Csin%28%5Ctheta%29)
The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.
Newton’s Second Law states that
![F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma%20%3D%20mg%5Csin%28%5Ctheta%29%5C%5C%5Csin%28%5Ctheta%29%20%3D%20a%2Fg)
Also, the car is making a circular motion:
![a = \frac{v^2}{R}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%5E2%7D%7BR%7D)
Combining the equations:
![\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta%29%20%3D%20%5Cfrac%7Ba%7D%7Bg%7D%20%3D%20%5Cfrac%7Bv%5E2%2FR%7D%7Bg%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7BgR%7D)
Finally the angle is
![\arcsin(v^2/(gR))](https://tex.z-dn.net/?f=%5Carcsin%28v%5E2%2F%28gR%29%29)