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nikdorinn [45]
3 years ago
10

Which device records digital signals? A. V C R tape. B. C D. C. phonograph record. D. audio cassette.

Physics
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

CD

Explanation:

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A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the
Rudik [331]

Answer:

Q must be placed at 0.53 L

Explanation:

Given  data:

q_1 = 4.0 μC , q_2 = 3.0μC

Distance between charge is L

third charge q be placed at  distance x cm from q1

The force by charge q_1 due to q is

F1 = \frac{k q q_1}{x^2}

F1 = \frac{k q ( 4.0 μC )}{ x^2}                  ----1

The force by charge q_2 due to q is

F2 =  \frac{k q q_2}{(L-x)^2}

F2 = \frac{kq (3.0 μC)}{(L-x)^2}                   --2

we know that net electric force is equal to zero

F_1 = F_2

\frac{k q ( 4.0 μC )}{x^2}   =\frac{k q ( 3.0 μC )}{(l-x)^2}

\frac{4}{3}*(L-x)^2 = x^2

x = \sqrt{\frac{4}{3}*(L - x)

L-x = \frac{x}{1.15}

L = x + \frac{x}{1.15} = 1.86 x

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Q must be placed at 0.53 L

3 0
3 years ago
Yeast breaking down starch into usable energy ____________ law of thermodynamics.
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4 0
3 years ago
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A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

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F = 4094.1 / 11

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4 0
3 years ago
The SI unit of power is the
fgiga [73]
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3 0
3 years ago
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Partially correct answer iconYour answer is partially correct. A proton initially has and then 2.30 s later has (in meters per s
morpeh [17]

Answer: The question has some missing details. The initial velocity given as u = -6.5i + 17j + 13k and the final velocity v = -2.8i + 17j -9.3k.

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b) magnitude = 9.85m/s2

c) direction = 280.64 degree

Explanation:

The detailed and step is shown in the attachment.

7 0
3 years ago
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