Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides
i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by
(b)
Magnetic flux,
(c)
R = 3 ohm
magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
Answer:
Distance will be 49.34 m
Explanation:
We have given wavelength
Diameter of the antenna d = 2.7 m
Range L = 7.8 km = 7800 m
We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D
We know that distance is given by
So distance D will be 49.34 m
Answer:
Explanation:
weight on moon = 1/6* weight on earth
69.3=1/6*weight on earth
weight on earth = 69.3*6
weight on earth = 415.8 N
Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m