Answer: 0.091 m
Explanation:
r = 1/B * √(2mV/e), where
r = radius of their circular path
B = magnitude of magnetic field = 1.29 T
m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg
V = potential difference = 2.9 kV
e = charge of the Uranium -238 ion = 1.6*10^-19 C
r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]
r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)
r = 1/1.29 * √0.0138
r = 1/1.29 * 0.117
r = 0.091 m
Therefore, the radius of their circular path is 0.091 m
The car's velocity is (distance + direction) / (time) =
(75 km-north) / (1.5 hrs) =
(75/1.5) (km-north/hr) = 50 km/hr north.
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
<span>Pice=920kg/m^3
deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa
P=Po + deltaP=101.325 + 9022 = 9123kPa</span>
Choice-C is a correct statement.
