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2 years ago

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A 15. 0 cm object is 12. 0 cm from a convex mirror that has a focal length of -6. 0 cm. What is the height of the image produced

by the mirror? –5. 0 cm 7. 5 cm -7. 5 cm 5. 0 cm.

1 answer:

ipn [44]2 years ago

6 0

Answer:

–5. 0 cm

Explanation:

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Explanation:

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A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc

Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s, u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest), = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

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torisob [31]

Ans: Intensity = I = $5.8 * 10^{-8} W/m^2$

Explanation:
First you need to find out the speed of Electromagnetic wave:

Since
v = $\sqrt{ \frac{1}{\mu\epsilon} }$

v = $\sqrt{ \frac{1}{\mu_r\mu_ok\epsilon_o} }$
$\mu_r = 5.18 \\
\mu_o = 4 \pi * 10^{-7} \\
k = 3.64 \\
\epsilon_o = 8.85 * 10^{-12}$

Plug in the values:
v = $\sqrt{ \frac{1}{(5.18)(4\pi*10^{-7})(3.64)(8.85*10^{-12})} }$
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Now that we have "v", we can use the following formula to find the intensity of wave:

$I = \frac{k\epsilon_o*v*E_{max}^2}{2}$

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