Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.
a) 1.57 m/s
The sock spins once every 2.0 seconds, so its period is
T = 2.0 s
Therefore, the angular velocity of the sock is
The linear speed of the sock is given by
where
is the angular velocity
r = 0.50 m is the radius of the circular path of the sock
Substituting, we find:
B) Faster
In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:
At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:
Therefore, the new linear speed would be:
And substituting,
So, we see that the linear speed has doubled.
Because the electrons collide with the particles inside the conductor so are therefore slowed down seen as current is the rate of flow of electrons
The 3003 aluminum alloy is made up of 1.25% Magnesium and 0.1% Copper. This combination is designed to increase the strength of the material over other types of alloys such as those of the 1000 series. This alloy provides a medium strength and can be educated by cold work.
The alloy is not heat treatable and generally has good formability, corrosion resistance and weldability.
However, being a material that hardens by cold work, welding a 3003 Aluminum structure will cause the body to undergo recrystallization which will generate a loss in the 'resistance' of the material and the force capable of withstanding. If this aluminum will be used for structural purposes, it should not be welded. It would be better to perform the structure with a 6061 aluminum, which has similar characteristics and is not so affected by welding.
Answer:
8.37×10⁻⁴ N/C
Explanation:
Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.
From the question, the expression for electric field is given as,
E = F/Q.......................... Equation 1
Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.
Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs
Substitute these values into equation 1
E = 8.2×10⁻² /98
E = 8.37×10⁻⁴ N/C
Hence the Electric Field of the charged balloon = 8.37×10⁻⁴ N/C
