What unbalanced force is needed to give a 976 kg vehicle an acceleration of 2.50 m/s2? ASAP
1 answer:
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it can be said that the speed of the east wind is
v=0.3608m/s
From the question we are told
A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).
After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,
- what is the speed of the east wind?.
Generally the equation for the distance is mathematically given as
BA=3000sin60
BA=2598.07m
Therefore
the speed of the east wind
v=0.3608
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Answer:
The number of turns, N = 1750
Explanation:
It is given that,
The inner radius of a toroid, r = 12 cm
Outer radius, r' = 15 cm
The magnetic field at points within the coils 14 cm from its center is,
R = 14 cm = 0.14 m
Current, I = 1.5 A
The formula for the magnetic field at some distance from its center is given by :
N = 1750
So, the number of turns must have in a toroidal solenoid is 1750. Hence, this is the required solution.