Answer:d
Explanation:
All the given situations are possible because
(a)When particles are oppositely charged then they attract each other
(b)One is Positively charged and other is uncharged: Charged particle will induce charges of opposite nature to attract the other particle
(c)Negatively charged particles will induce the positive charge in the uncharged particle to attract the initially uncharged particle.
Answer:
Hello your question is incomplete attached below is the complete question and solution
<em>answer; </em>
attached below
Explanation:
<em>Given data:</em>
100-turn coil
mean length of central leg = 5.5 cm
mean length of outer paths = 15.5 cm
relative permeability = 2000
cross sectional area ( A ) = 1 cm^2
distance x = 1 cm
attached below is a detailed solution
Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.

where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
![A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }](https://tex.z-dn.net/?f=A%20%3D%204%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B100%5E%7B2%7D%20cm%5E%7B2%7D%20%7D)
A = 0.0004 [m²]
Also, the weight should be calculated as follows:

where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
![w = 60*9.81\\w = 588.6[N]](https://tex.z-dn.net/?f=w%20%3D%2060%2A9.81%5C%5Cw%20%3D%20588.6%5BN%5D)
And the pressure:
![P=588.6/0.0004\\P=1471500 [Pa]](https://tex.z-dn.net/?f=P%3D588.6%2F0.0004%5C%5CP%3D1471500%20%5BPa%5D)
Because 1 [Pa] = 1 [N/m²]
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V