2 minutes = 120 seconds
120/15 = 8
The black horse corresponds to 8 seconds.
Answer: The result to the query is 0.25 ohms
Explanation:
R = u x/A.......1
where u represents the resistivity of the
rod, A is the cross-sectional area, and x denotes
the length of the rod.
Let R* represent the resistance across the adjacent sections of the rod.
Then, R* = u1/4.......2
By comparing equation 1 with equation 2, we find that
R* = 1/4
which equals 0.25 ohms.
We will express each number differently:
(4.48E-8) = 0.0000000448
(5.2E-4) = 0.00052
Upon performing the multiplication, the exponent will align to 10 ^ -11
Consequently, the multiplication yields:
(4.48E-8) * (5.2E-4) = 2.3296E-11
Rewriting gives us:
(4.48E-8) * (5.2E-4) = 2.33E-11
Response:
2.33E-11
A block weighing m1 = 4.50 kg and a ball weighing m2 = 7.70 kg are linked by a light string over a frictionless pulley, as illustrated in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300.
(a) Determine the acceleration of both objects and the tension present in the string.
(b) Verify the acceleration calculation using a systems approach. (Utilize m1, m2, μk, and g as needed.)
(c) If an additional mass is added to the ball, what amount is necessary to augment the downward acceleration by 60%?
In (a), I calculated the acceleration to be 5.10 m/s^2 and the tension to be 36.2N.
In (b), the equation used is a = (m2g-ukm1g)/(m1+m2)
It’s (c) that I am struggling to understand. Can anyone assist me?
Answer:
Explanation:
Given,
Voltage of the primary coil (V_p) = 30 kV-rms
Voltage of the secondary coils (V_s) = 345 kV-rms
number of turns in the primary coil (n_p) = 80 turns
number of turns in the secondary coil (n_s) =?
the ratio of turns between primary and secondary coils
The number of turns in the secondary coil is equal to
