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Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;
Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
Therefore, the extension of the wire is 0.362 mm.
700 makes the maximum output power.
<u>Explanation:</u>
In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.
A joule is equal to one Newton-meter, which is the amount of work needed to move a 1 Newton force a distance of 1 meter. When you divide work by time, you get power, measured in units of joules per second. This is also called a Watt. 1 Watt = 1 Joule Sec. This is the formula to calculate output power.
Hookes law state that provided that the elastic limit is not exceeded, the extension is directly proportional to the force
Uniform velocity means no Net force and therefore no acceleration. Acceleration only happens when the velocity changes.