Answer:
a) 9.167
b) 402.5 J
Explanation:
We have a machine, which when applying a force of 60N, it returns 550N.
a) The mechanical advantage of a machine is the number of times that the machine increases an input force.
Mathematically, is the relationship between the output force and the input force.
The mechanical advantage is dimensionless.
In this case, the input force (the force we apply) is 60 N; and the output force (the force that the compound machine exerts) is 550 N.
We solve it this way:
This is the mechanical advantage of the machine in this exercise.
b)In a machine, we define the efficiency as the relationship between the work we receive from the machine, and the work we give to it.
The machine efficiency is a real number between 0 and 1 and is dimensionless.
In this exercise, the efficiency is 35% = 0.35
We solve it this way :
We obtain 402.5 J useful work from the machine.
Answer:
The correct answer to the following question will be "Period".
Explanation:
The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.
It is denoted by "T" and its formula will be:
⇒
Where, T = Period
F = Frequency
The other given choices are not related to the given circumstances. So that the above would be the right answer.
Answer:
<em>0.85c </em>
Explanation:
Rest mass of Kaon = 494 MeV/c²
Rest mass of proton = 938 MeV/c²
The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²
for the kaon, rest energy = 494c² MeV
for the proton, rest energy = 938c² MeV
Recall that the rest energy, and the total energy are related by..
= γ
which can be written in this case as
= γ
...... equ 1
where = total energy of the kaon, and
= rest energy of the kaon
γ = relativistic factor =
where
But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...
=
......equ 2
where is the total energy of the kaon, and
is the rest energy of the proton.
From =
= 938c²
equ 1 becomes
938c² = γ494c²
γ = 938c²/494c² = 1.89
γ = = 1.89
1.89 = 1
squaring both sides, we get
3.57( 1 - ) = 1
3.57 - 3.57 = 1
2.57 = 3.57
= 2.57/3.57 = 0.72
= 0.85
but,
v/c = 0.85
v = <em>0.85c </em>