Frequency = 1 / (period)
Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.
Answer:
(4) 8.5 m/s
Explanation:
You add both the meters together and both the seconds together and then divide them both.
Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.
r^² x Pressure (pa).
Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.
Explanation:
Answer:
Explanation:
For resistance of a wire , the formula is as follows
R = ρ L / S
where ρ is specific resistance , L is length and S is cross sectional area
Given L = 14 000 m ,
S = 4.8 x 10⁻⁴ m²
specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter
Putting the values in the formula
R = 2.8 x 10⁻⁸ x 14 x 10³ / (4.8 x 10⁻⁴ )
R = 0.8167 ohm .
= .82 ohm .
