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3 years ago

What is the correct answer?

Physics

1 answer:

LuckyWell [14K]3 years ago

7 0

Answer:

Option (4)

Explanation:

Given polynomial is,

P(x) = 1x³ - 7x² + 2x + 40

If (x - 5) is a factor of the given polynomial,

Synthetic division by x = 5,

5 | 1 -7 2 40

1 -2 -8 0

Therefore, factored form of the polynomial will be,

P(x) = (x - 5)(x²- 2x - 8)

Now we further factorize the expression (x² - 2x - 8),

(x² - 2x - 8) = x²- 4x + 2x - 8

= x(x - 4) + 2(x - 4)

= (x + 2)(x - 4)

Now fully factored form of the polynomial will be,

P(x) = (x - 5)(x + 2)(x - 4)

Therefore, remaining factors of the polynomial other than (x - 5) are (x - 4)(x + 2)

Option (4) will be the correct option.

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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00

Basile [38]

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

$V=initial\ rate\times inductance$

$initial\ rate=\dfrac{V}{L}$

Put the value into the formula

$initial \rate=\dfrac{6.00}{2.30}$

$initial\ rate=2.60\ A/s$

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

$rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}$ ....(I)

Where, $T=\dfrac{L}{R}$

$T=\dfrac{2.30}{8.00}$

$T=0.2875$

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

$e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})$

$e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})$

$e^{\dfrac{-t}{T}}=0.333$

The rate of current increases is

Put the value in the equation (I)

$rate\ of\ current\ increase=2.60\times0.333$

$rate\ of\ current\ increase=0.8658\ A/s$

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

Put the value into the formula

$i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})$

$i=0.435\ A$

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

$i=\dfrac{V}{R}$

$i=\dfrac{6.00}{8.00}$

$i=0.75\ A$

The final steady-state current is 0.75 A.

Hence, This is the required solution.

3 0

3 years ago

An object with little mass won't require a lot of force to move.

Anna35 [415]

Answer:

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3 years ago

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Answer:

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4 years ago

At a construction site, the site manager notices that a crane takes 20 seconds to lift a 500kg steel beam up to a height of 15 m

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4 years ago