We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
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<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
Answer:
Speed of gamma rays = 3 x 10⁸ m/s
Explanation:
Given:
Frequency of gamma ray = 3 x 10¹⁹ Hz
Wavelength of gamma rays = 1 x 10⁻¹¹ meter
Find:
Speed of gamma rays
Computation:
Velocity = Frequency x wavelength
Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays
Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]
Speed of gamma rays = 3 x [10¹⁹⁻¹¹]
Speed of gamma rays = 3 x [10⁸]
Speed of gamma rays = 3 x 10⁸ m/s
Answer:
Part a)
V = 18.16 V
Part b)
Part c)
P = 672 Watt
Part d)
V = 5.84 V
Part e)
Explanation:
Part a)
When battery is in charging mode
then the potential difference at the terminal of the cell is more than its EMF and it is given as
here we have
now we have
Part b)
Rate of energy dissipation inside the battery is the energy across internal resistance
so it is given as
Part c)
Rate of energy conversion into EMF is given as
Now battery is giving current to other circuit so now it is discharging
now we have
Part d)
Part e)
now the rate of energy dissipation is given as
