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1 year ago

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Use the formula from Newtons second law, namely Fnet=∆t=∆P that the Net force exerted on the object is equal to the product of t

he mass and acceleration of the object

1 answer:

Aleksandr-060686 [28]1 year ago

6 0

Answer:

Explanation:

Fnet = Δt = ΔP

Fnet = ΔP/Δt

= mv - mu / t

= m(v - u) / t

But we know v = u + at

=> a = v - u / t

Therefore,

Fnet = ma

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Explanation:

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Where;:

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Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

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When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

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$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

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