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Nitella [24]
3 years ago
8

Which mass is undergoing to the greatest amount of acceleration ??

Physics
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

Option (3)

Explanation:

Formula used to calculate acceleration is,

F = ma

Where F = force exerted on a mass

m = mass

a = acceleration due to force exerted on the mass

Option (1),

When F = 100 N and m = 100 kg

100 = 100a

a = 1 m per sec²

Option (2)

For F = 1 N and m = 100 kg

1 = 100a

a = \frac{1}{100}

a = 0.01 m per sec²

Option (3)

For F = 100 N and m = 1 kg

100 = 1(a)

a = 100 m per sec²

Option (4)

For F = 1 N and m = 1 kg

1 = 1(a)

a = 1 m per sec²

Therefore. acceleration in Option (3) is the maximum.

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E_1 =- \frac{13.6 }{1^2}

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Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

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Wavelength in nm = 1244 / energy in eV

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A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the
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Answer:

1.8 cm

Explanation:

m = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

q = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

\Delta V =Potential difference through which the ion is accelerated = 215 V

v = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}

r = Radius of the path followed by ion

B = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm

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