I think this is the answer. I hope you can understand.
Answer:
F = 0i (in the x-direction), 0j (in the y-direction),-8.59*10^-4 N k (In the z-direction)
Explanation:
The force given by charged particles moving in a magnetic field is given below (cross is cross product, they don't have that format in the equation tool):
Now we can perform the cross product between v and B
d{array}\right][/tex]
Now multiply by Q (charge) to get the force
F = -8.59*10^-4 N k
F = 0i, 0j, (-8.59*10^-4)k
Explanation:
The initial kinetic energy is
The final kinetic energy is
The work done W on the car is
The power expended P is
<span>The answers are --
a) wind direction
b) wind speed
e) intensity of precipitation
f) location of precipitation</span>
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s