Answer:
Explanation:
A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.
In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.
When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.
Answer:


Explanation:
what is the smallest crater that each of these telescopes could resolve on our moon?
For moon ;
s = 3.8 × 10 ⁸ m
y = 1.22 λs/D
where;
λ = 400 nm = 400× 10 ⁻⁹
D = 2.4 m
The smallest crater for the hubble space is calculated as follows:


For Aceribo ;
y = 1.22 λs/D
where :
λ = 75 cm = 0.75 m
D = 305 m


Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :


W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
14-needle heading west
15-the strength of the current and the distance