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Andreyy89
3 years ago
7

A charge of 8.4*10^-4 moves at an angle of 35 degrees to a magnetic field that has a field strentgh of 6.7

Physics
1 answer:
12345 [234]3 years ago
3 0

The force on a charged particle in a magnetic field is given by

the speed of the charged particle = 10842 m/s.

Explanation:

F= q V B sinθ

F=force=3.5 x 10⁻²N

q= charge= 8.4 x 10⁻⁴ C

B= magnetic field= 6.7 x 10⁻³ T

θ=35⁰

Thus the velocity is given by V=\frac{F}{q B sin35}

V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]

V=10842 m/s

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4 years ago
Prove that..<br>please help<br>​
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\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

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❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

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\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

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<u>━━━━━━━━━━━━━━━━━━━━</u>

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