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Answer:
It is 19.6 m below the helicopter.
Explanation:
Given that,
Initial velocity = 2 m/s
Time t = 2 sec
(I). The velocity of the packet is
Using equation of motion
Negative sign shows that the direction of downward.
(II)The distance covers by the helicopter is
The distance covers by the packet
Using equation of motion
The total distance is
Hence, It is 19.6 m below the helicopter.
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Sia ? 6 years, 6 months ago
- For a.c. voltage
{tex}V = {V_0}\;\sin \left( {\omega t - \phi } \right){/tex}
The peak value
{tex}{V_0} = 220\sqrt 2 = 311V{/tex}
The rms value of voltage
{tex}{V_{rms}} = \frac{{{V_0}}}{{\sqrt 2 }};\;{V_{rms}} = 220V{/tex} - Average voltage in full cycle is zero. Average voltage in half cycle is
{tex}{V_{av}} = \frac{2}{\pi }{V_0} = \frac{2}{\pi } \times 311 = 198.17V{/tex} = 198.17 V - As {tex}\omega = 2\pi f,\;2\pi f = 314{/tex}
i.e. {tex}f = \frac{{314}}{{2 \times \pi }} = 50\;Hz{/tex} = 50 Hz
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You can check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html
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Decay constant of a radioactive element is the reciprocal of the time during which the number of atoms left in the sample reduces to {tex}\frac{1}{e}{/tex} times the original number of atoms in the sample.
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According to Rutherford model, there is a positively charged centre in an atom called the nucleus and the electron present in nucleus revolve aroun it, in well- defined orbits. But, since an electron is charged particle. Thus, while revolving around the nucleus it lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.
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Riti Bansal 6 years, 6 months ago
1Thank You