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3 years ago

9

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

is needed to select a speed of 4.40 ✕ 106 m/s? V/m

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

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According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

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Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

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The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

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