Answer:
Explanation:
The force exerted in a magnetic field is given as
F = q (v × B)
Where
F is the force entered
q is the charge
v is the velocity
B is the magnetic field
Given that,
The magnetic field is
B = 2•i + 4•j. T
The velocity of the electron is
v = 2•i + 6•j + 8•k. m/s
Also, the charge of an electron is
q = -1.602 × 10^-19 C.
Then note that,
V×B is the cross product of the speed and the magnetic field
Then,
F = q (V×B)
F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)
Note
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]
F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]
F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]
F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)
F = -1.602 × 10^-19(16•j - 32•i)
F = -1.602 × 10^-19 × ( -32•i + 16•j)
F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j
Then, the x component of the force is
Fx = 5.126 × 10^-18 N
Also, the y component of the force is
Fy = -2.563 × 10^-18 N
Voltage (V)= Current (I) * Resistance (R)
I=V/R=140/2=70A
Hope this helps!
Answer:
Since the distance between the pivot and its weight is zero, the moment of its weight about the pivot (= weight × 0) is zero. Hence, the weight of the ruler can be ignored.
Hope this helps:)
This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
<em />
<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
Answer:
Explanation:
Given
two holes are made with different sizes
Hole 1 is large in size and hole 2 is small
If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.
Now for radius apply Bernoulli's theorem at hole 1 and 2


if hole 1 is h distance below water surface then 
and 
Also 

and 

thus 

