Answer:
I = 0.002593 A = 2.593 mA
Explanation:
Current density = J = (3.00 × 10⁸)r² = Br²
B = (3.00 × 10⁸) (for ease of calculations)
The current through outer section is given by
I = ∫ J dA
The elemental Area for the wire,
dA = 2πr dr
I = ∫ Br² (2πr dr)
I = ∫ 2Bπ r³ dr
I = 2Bπ ∫ r³ dr
I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]
I = (Bπ/2) [R⁴ - (0.9R)⁴]
I = (Bπ/2) [R⁴ - 0.6561R⁴]
I = (Bπ/2) (0.3439R⁴)
I = (Bπ) (0.17195R⁴)
Recall B = (3.00 × 10⁸)
R = 2.00 mm = 0.002 m
I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]
I = 0.0025929449 A = 0.002593 A = 2.593 mA
Hope this Helps!!!
Answer:
3600 J
Explanation:
According to given question
P(rated)=60w
V= 120
I =0.50 A
t=600 second
Now,
Energy can be calculated as :
Where,
V is voltage
I is current
t is time in second
Now,
Putting the all value in above equation E
So,
Therefore, 36000 J energy use up by light bulb
Answer:
★The second law of refraction
The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant
This constant value is called the refractive index of the second medium with respect to the first.
Answer:
I believe it's frictional force
The frontal bone.
A blowout fracture is one where the orbit of the eye socket is damaged. This type of fracture is usually accompanied by double vision, depressed ocular globes and loss of feeling in the cheeks. Bones other than the frontal bone which are responsible for the structure of the eye socket are the zygomatic bone and the maxilla.
