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3 years ago

Material speed of light

2 answers:

3 0

The question is poor. Light doesn't refract on its way THROUGH anything. It refracts at the boundary BETWEEN two different media. The effect is greatest where the ratio of the speeds of light in the two media is greatest. On your list, that would be at the boundary between air or space and glass.

saul85 [17]3 years ago

3 0

Answer: I think the answer is B glass.

Explanation: Because Glass is light.

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Find the resistance of an electric light bulb if a current of 0.08 A flows when the potential difference across the bulb is 120

lora16 [44]

U=RI Ohm's law
then R=U/I
=120/0.08
=2250Ω
hope this helps you

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3 years ago

What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7

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<h3>What does the Earth's center's gravitational pull feel like?</h3>

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<h3>Where is the Earth's and the moon's gravitational centre?</h3>

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6 0

1 year ago

Which of the following is not a part of dalton s atomic theory?

Alex777 [14]

<span>c. atoms are always in motion..............</span>

6 0

3 years ago

Read 2 more answers

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago