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3 years ago

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create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

5 words and 10 should come from your book. These do not have to rhyme, but can if you wish.

1 answer:

Luba_88 [7]3 years ago

8 0

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

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A weight lifter does 586 J of work on a weight that he lifts in 3.5 seconds. What is the power with which he lifts the weight?

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Answer:

Explanation:

The quantity of energy transferred by a force when it is applied to a body and causes that body to move in the direction of the force work.

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2 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

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3 years ago

Which of the following shows the prefixes in order from smallest to largest?

Ulleksa [173]

A) micro milli centi

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3 years ago

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A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

belka [17]

We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values,
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶) / (3 * 10⁻²)²

F = 100 N [ East of positive charge ]

Hope this helps!

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3 years ago

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A balloon is charged until it has a voltage of v=9800V. A student reaches out and is shocked by a spark. The electric force does

alina1380 [7]

a)]An expression for the magnitude of charge transferred Q is Q = W/V.

b) The magnitude of the charge, in coulombs is 1.02 x 10⁻⁹ Coulomb.

c) Number of electrons in this are 63.75 x 10⁸ electrons.

<h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

A balloon is charged until it has a voltage of v=9800V. A student reaches out and is shocked by a spark. The electric force does W= 1x10⁻⁶ J of work transferring the charge.

a) Work done = Charge x potential

W = Q x V

Q = W/V

b)Substitute the values into the expression, we have

|Q| = 1x10⁻⁶/9800

|Q| = 1.02 x 10⁻⁹ Coulomb

c) No of electrons = total charge /charge on electron

n = 1.02 x 10⁻⁹/1.6 x 10⁻¹⁹

n = 63.75 x 10⁸ electrons

Learn more about electrons.

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2 years ago