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2 years ago

What is the strength of an electric field that will put a force of 1.28 x 10-15 N on a proton?

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

8 0

Answer: E = 7,490.6 N/C

Explanation:

If we have a field E, and a particle with a charge q, the force that the particle experiences is:

F = E*q

In this case, we know that the force is:

F = 1.2*10^(-15) N

And we know that the particle is a proton, where the charge of a proton is:

q = 1.602*10^(-19) C

Then we can replace these two values in the equation to get:

1.2*10^(-15) N = E*1.602*10^(-19) C

We just need to isolate E.

(1.2*10^(-15) N)/(1.602*10^(-19) C) = E

7,490.6 N/C = E

That is the strength of the electric field.

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If a string completes 500 vibrations in one second, what is its frequency?

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Answer:

500 Hz

Explanation:

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3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

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3 years ago

What are some ways houses along the coastlines can protect themselves from storm surges?

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Build walls around the coast

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3 years ago

An example of forced convection is the movement of hot air by a fan. true or false?

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The answer is true. The blowing fan force the hot air to go through the process of convection .

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3 years ago

Read 2 more answers

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

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3 years ago