1.8 is the mechanical advantage of the lever.
<h3>Definition of mechanical advantage</h3>
The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.
The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.
Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.
In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.
Mechanical advantage= 90/50 =1.8
Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?
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It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave. Radio and light are both electromagnetic waves. The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have. But they're both the same physical phenomenon.
However, a radio wave CAN'T also be considered to be a sound wave. These two things are as different as two waves can be.
-- Radio is an electromagnetic wave. Sound is a mechanical wave.
-- Radio waves travel more than 800 thousand times faster than sound waves do.
-- Radio waves are transverse waves. Sound waves are longitudinal waves.
-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.
-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.
-- Sound waves can be detected by our ears. Radio waves can't.
-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.
-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.
Answer:
(a) the mechanical energy of the system, U = 0.1078 J
(b) the maximum speed of the object, Vmax = 0.657 m/s
(c) the maximum acceleration of the object, a_max = 15.4 m/s²
Explanation:
Given;
Amplitude of the spring, A = 2.8 cm = 0.028 m
Spring constant, K = 275 N/m
Mass of object, m = 0.5 kg
(a) the mechanical energy of the system
This is the potential energy of the system, U = ¹/₂KA²
U = ¹/₂ (275)(0.028)²
U = 0.1078 J
(b) the maximum speed of the object
(c) the maximum acceleration of the object
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done
The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.
The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy
The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
<h2>It solved by the Hooke's law states F=kx</h2>
answer is
<h2>0.4n/m</h2>
