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2 years ago

What is magnetic field right and left hand rule

Physics

1 answer:

baherus [9]2 years ago

7 0

Answer:

The Right Hand Rule, illustrated at left, simply shows how a current-carrying wire generates a magnetic field. ... The Left Hand Rule shows what happens when an electrical current enters a magnetic field. You need to contort your hand in an unnatural position for this rule, illustrated below

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Elements of Music Timbre Dynamics Rhythm Form Texture Harmony Style

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Explanation:

Rhythm, Harmony, Timbre, Dynamics, Texture, and Form

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2 years ago

in 1859 a hunter brought 24 rabbits from England to Australia and release them to establish a population for sport hunting rabbi

slavikrds [6]

I would say the rabbits were breeding.

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3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

2 years ago

Consider the circuit below, which is powered by a 8-v battery. switch s is opened at t = 0 after having been closed for a long t

GrogVix [38]

The battery will be full still a 8v bc of no time comparison

4 0

3 years ago

Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024

myrzilka [38]

Given: Mass of earth Me = 5.98 x 10²⁴ Kg

Radius of earth r = 6.37 x 10⁶ m

G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r

but V = 2πr/T

equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²

GMe = 4π²r³/T²

T² = 4π²r³/GMe

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds or around 1.4 Hour

3 0

2 years ago

What is magnetic field right and left hand rule​

What is magnetic field right and left hand rule