I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
Answer:
U = 25 J
Explanation:
The energy in a set of charges is given by
U = 
in this case we have three charges of equal magnitude
q = q₁ = q₂ = q₃
with the configuration of an equilateral triangle all distances are worth
d = a
U = k (
)
we substitute
15 = k q² (3 / a)
k q² /a = 5
For the second configuration a load is moved to the measured point of the other two
d₁₃ = a
The distance to charge 2 that is at the midpoint of the other two is
d₁₂ = d₂₃ = a / 2
U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})
U = k q² (
)
U = (kq² /a) 5
substituting
U = 5 5
U = 25 J
Answer
given,
Mass of Kara's car = 1300 Kg
moving with speed = 11 m/s
time taken to stop = 0.14 s
final velocity = 0 m/s
distance between Lisa ford and Kara's car = 30 m
a) change in momentum of Kara's car
Δ P = m Δ v


Δ P = - 1.43 x 10⁴ kg.m/s
b) impulse is equal to change in momentum of the car
I = - 1.43 x 10⁴ kg.m/s
c) magnitude of force experienced by Kara
I = F x t
I is impulse acting on the car
t is time
- 1.43 x 10⁴= F x 0.14
F = -1.021 x 10⁵ N
negative sign represents the direction of force
Answer:
The magnitude will be "
". The further explanation is given below.
Explanation:
Trying to determine what is usual for a rectangular loop plane,
⇒ 
Magnetic moment is given as:
μ = 
On putting the values in the above formula, we get
μ = ![(5.5A)[(0.3m)(0.4m)][(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(-\hat{k})]](https://tex.z-dn.net/?f=%285.5A%29%5B%280.3m%29%280.4m%29%5D%5B%28Cos35%5E%7B%5Ccirc%7D%29%28%5Chat%7Bi%7D%29%2B%28Sin35%5E%7B%5Ccirc%7D%29%28-%5Chat%7Bk%7D%29%5D)
= ![0.66 Am^2[(Cos35^{\circ})(\hat{i})+(Sin35^{\circ})(\hat{k})]](https://tex.z-dn.net/?f=0.66%20Am%5E2%5B%28Cos35%5E%7B%5Ccirc%7D%29%28%5Chat%7Bi%7D%29%2B%28Sin35%5E%7B%5Ccirc%7D%29%28%5Chat%7Bk%7D%29%5D)
Now the external vector-shaped torque seems to be:
Magnitude,

On putting the values in the above formula, we get
⇒ ![\sigma=[(0.06Am^2)(Cos35^{\circ}(-\hat{i})+Sin35^{\circ}(-\hat{k})]\times (2.9T)(-\hat{i})](https://tex.z-dn.net/?f=%5Csigma%3D%5B%280.06Am%5E2%29%28Cos35%5E%7B%5Ccirc%7D%28-%5Chat%7Bi%7D%29%2BSin35%5E%7B%5Ccirc%7D%28-%5Chat%7Bk%7D%29%5D%5Ctimes%20%282.9T%29%28-%5Chat%7Bi%7D%29)
⇒ 
⇒ 
⇒ 