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ratelena [41]
3 years ago
9

Rebecca forgets to title her notes in her physics

Physics
2 answers:
kolbaska11 [484]3 years ago
6 0

Answer:

use of nanotechnology ; quantum dots

Explanation:

edg 2021

choli [55]3 years ago
4 0

Answer:

Explanation:

The notes are about using Quantum Dots for bio-medical imagining. The notes are more on its uses than its theory. So the title should be O Uses of Nanotechnology: Quantum Dots.

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A printed circuit board (PCB) is supported by a chassis that is attached to a vibrating motor. The board is 1.6 mm thick, 200 mm
Vedmedyk [2.9K]

jtejyrkekdkeludmydmumud

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3 years ago
the average speed of a runner in a 483. meter race is 3.0 meters per second. How long me runner to complete the race? Dont inclu
lara [203]

Answer:

161

Explanation:

v=\frac{d}{t} slove for t

t=\frac{d}{v}

Insert values of d and v

t=\frac{483}{3} \\

t=161

3 0
3 years ago
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?
prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about 

                     (1 / √0.66) = 1.23 times

the Earth's radius, so the height is about  910 miles above the surface.


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3 years ago
Form conise note on heat energy specific heat application evaporation boiling sublimation relative humidity and dew point
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The answer is letter a

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