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mash [69]
4 years ago
7

a beam of electrons passes through an electric field where the magnitude of the electric field strength is 3.00 * 10^3 newtons p

er coulomb. what is the magnitude of the electrostatic force exerted by the electric field on each electron in the beam?
Physics
1 answer:
nikdorinn [45]4 years ago
5 0
Electric Field Strength E = Force F/ Charge q

E = F/q

Force , F =   E*q

But charge on an electron q = 1.6 * 10⁻¹⁹ C

E = 3 * 10³ N/C from the question.

Force F =  3* 10³ *   1.6 * 10⁻¹⁹

F = 4.8 * 10⁻¹⁶  N.

Electrostatic force F = 4.8 * 10⁻¹⁶  N
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Answer:

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5 0
2 years ago
Read 2 more answers
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Snezhnost [94]

Explanation:

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B(t)=B_oe^{-\dfrac{t}{\tau}}

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\epsilon=\dfrac{-d\phi}{dt}

Where

\phi=BA

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\epsilon=NAB_o\dfrac{-d(e^{-\dfrac{t}{\tau}})}{dt}

\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}

Power dissipated is given by :

P=\dfrac{\epsilon^2}{R}

P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}

P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}

Energy dissipated in the circuit is given by :

E=\int\limits^T_0 {P.dt}

E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}

E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}

E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}

Hence, this is the required solution.

7 0
3 years ago
A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

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Substituting

           F = kx

           5.3 = k x 0.36

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Spring constant is 14.72 N/m

6 0
3 years ago
After being pushed and released, a 50 kg crate slides across a factory floor. Friction on the sliding crate is 200 N, what is th
saveliy_v [14]
F = ma
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6 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
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