When plane is flying along the wind then we can say
Now when its going against the wind the speed is given by
Now by the above two equations we will have
NOooOoIooOOi Hsushrhndndnrbrbr rgehrbhrhrbr r he rjrnd
The unit of the quotient of inductance and resistance will be Henry and ohm
Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. The flow of electric current creates a magnetic field around the conductor. The field strength depends on the magnitude of the current, and follows any changes in current.
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
unit of quotient of inductance = henry (H)
unit of resistance = ohm
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The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
The momentum increases by a factor of 2
Explanation:
We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.
The kinetic energy of the rocket is:
(1)
where
m is the mass
v is the velocity
The momentum of the rocket is
(2)
From eq.(1) we get
and substituting into (2),
Now in this problem we have:
- The kinetic energy of the rocket is increased by a factor 8:
- The mass is reduced by half:
Substituting, we find the new momentum:
So, the momentum increases by a factor of 2.
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