1 point energy should be the answer
Answer:
f1/f2 =W1/W2 = 1/3
.0 f2 = 3f1
As ,
1/F= 1/f1 +1/f2
...1/40 = 1/f1 - 1/3f1
f1=> 80/3 cm
... f2 = 2f1 = 3 x 80/3 = 80 cm
Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer:
Y
Explanation:
Y will produce the strongest magnetic force because the current intensity is medium and it has 50 turns in the wire. the number of turns will increase the force because the magnetic field will be more concentrated.
Have a great day :)