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3 years ago

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A disk with a rotational inertia of 2.5 kg-m2 and a radius 1.1 m rotates on a frictionless fixed axis perpendicular to the disk

faces and through its center. A force of 7.7 N is applied tangentially to the rim. The angular acceleration of the disk is _____ rad/s2. Round your answer to the nearest tenth.

1 answer:

soldier1979 [14.2K]3 years ago

7 0

Answer:

3.4 rad/sec^2

Explanation:

rotational inertia = 2.5 kg-m^2 radius = 1.1 m force = 7.7 N

t = rotational inertia * angular acceleration equation 1

also t = force * radius

therefore to calculate angular acceleration equation 1 becomes

f * r = inertia * angular acceleration hence

angular acceleration = f * r / inertia = $\frac{7.7 * 1.1 }{2.5}$ 8.47 / 2.5 = 3.388 ≈ 3.4 rad/sec^2

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An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then

seropon [69]

Answer:

The charge stored is $Q = 4.25 *10^{-7 } \ C$

The energy stored is $E = 2.55*10^{-6} \ J$

Explanation:

From the question we are told that

The area of the plates is $A = 0.40 \ m^2$

The separation between the plate is $d = 0.10 \ mm =0.0001 \ m$

The potential difference is $V = 12 \ V$

The permitivity of free space is $\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2$

The dielectric constant of glass is K = 5.0

Generally the capacitance of this capacitor is

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substituting values

$C = \frac{8.85*10^{-12} * 0.40}{0.0001}$

$C = 3.34 *10^{-8} \ C$

The charge stored is mathematically evaluated as

$Q = CV$

substituting values

$Q = (3.54*10^{-8} * 12)$

$Q = 4.25 *10^{-7 } \ C$

The energy stored is

$E = 0.5 * CV^2$

substituting values

$E = 0.5 * (4.25 *10^{-7} * 12^2)$

$E = 2.55*10^{-6} \ J$

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3 years ago

Which item is heavier, 50 kg of cotton or 50 kg of iron?

Kipish [7]

Answer:

They weight the same

Explanation:

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3 years ago

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

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$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

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3 years ago

A carpenter on the roof of a building accidentally drops herhammer. As the hammer falls it passes two windows of equal height,a)

nadezda [96]

Answer:

Explanation:

a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.

b) III

The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.

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3 years ago