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3 years ago

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Three magnets are placed on a plastic stick as shown in the image. Explain how the magnets need to be rearranged so that they st

ick together and why rearranging them this way allows them to stick together.

2 answers:

dimaraw [331]3 years ago

6 0

Answer:

The magnets need to be placed with red closest to blue.

Opposite poles attract.

The magnets will be attracted to each other with enough force to stick together.

Explanation:

svlad2 [7]3 years ago

4 0

Answer:

The magnets need to be arranged so that the blue side of one magnet faces the red side of another magnet. When they are arranged in this way, the opposite poles of the magnets will be closest together. Because opposite poles attract, the force of the magnets will make the magnets stick together.

Explanation:

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An object moves along the x-axis. Its position is given by the equation y( X = 4t^2 - 41t+ 78 \). Find the position of the

prohojiy [21]

The object changes direction when its velocity changes sign. You can get the velocity function by differentiating the position function with respect to time <em>t</em> :

<em>x(t)</em> = 4<em>t</em> ² - 41<em>t</em> + 78

→ <em>v(t)</em> = d<em>x(t)</em>/d<em>t</em> = 8<em>t</em> - 41

Solve <em>v(t)</em> = 0:

8<em>t</em> - 41 = 0

8<em>t</em> = 41

<em>t</em> = 41/8 = 5.125

Just to confirm that the velocity indeed changes sign:

• Pick any time before this one to check the sign of <em>v</em> :

<em>v</em> (0) = 8•0 - 41 = -41 < 0

• Pick any time after and check the sign again:

<em>v</em> (6) = 8•6 - 41 = 7 > 0

Now just find the position at this time:

<em>x</em> (5.125) = -433/16 = -27.0625

which means the object is 27.0625 units on the negative <em>x</em>-axis.

You can also do this without calculus by completing the square in the position function:

4<em>t</em> ² - 41<em>t</em> + 78 = 4 (<em>t</em> ² - 41/4 <em>t</em> ) + 78

… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)² - (41/8)²) + 78

… = 4 (<em>t</em> ² - 2• 41/8 <em>t</em> + (41/8)²) - 4•1681/64 + 78

… = 4 (<em>t</em> - 41/8)² - 433/16

which describes a parabola that opens upward. When <em>t</em> = 41/8 = 5.125, the quadratic term vanishes and the turning point of the parabola occurs at a position of -433/16 units.

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3 years ago

When you look at the light from distant stars you are really looking back in time explain what this means

Leona [35]

Answer: Stars are in space for very long time, much longer than that one night. You are looking back in time because those stars have been there for so long that it’s like looking back in time, to when those stars were there.

Explanation:

May I please have brainlest

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3 years ago

Read 2 more answers

1. A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass.

vredina [299]

A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass = <span>A greater effect.</span>

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3 years ago

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Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and

Lelu [443]

Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD, $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D $_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r $_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D $_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r $_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However $\phi_r = \phi_{C/D} - \phi_{B/A}$

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$ = Torque on shaft AB and CD respectively

$T_{AB}$ = Required

$T_{CD}$ = 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$ = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$ = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

= $\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T $_{AB}$ = 79.54 N.m

Where T = $T_{AB}$ + T $_{CD}$ =

Therefore T $_{CD total }$ = 500 - 79.54 = 420.46 N·m

τ $_{max}$ = $\frac{T\times R}{J}$

$\tau_{ABmax}$ = $\frac{T_{AB}\times R_{AB}}{J_{AB}}$ = $\frac{79.54\times 0.015}{7.95\times 10^{-8}}$ = 15 MPa

$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa

7 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

a)

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

b)

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago