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seraphim [82]
4 years ago
5

I have a project due in 3 days and I need help!

Physics
1 answer:
Flura [38]4 years ago
8 0
1.Phases - Gas, liquid and solid. Gas, liquid, and solid are known as the three states of matter or material, but each of solid and liquid states may exist in one or more forms.
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Plz do all plz i will give brainest and thanks to best answer do it right
lara31 [8.8K]

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I'm not really sure but I think it's choice a

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What does resonance result in?
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“The vibrations of the aluminum force the air Column inside of the rod to vibrate at its natural frequency. The match between the vibrations of the air column and one of the natural frequencies of the singing rod causes resonance.”
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3 years ago
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If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f
Novay_Z [31]

Answer:

E =  ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 \pi R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o   ......1

so first we find  Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( \pi R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 \pi R × l)  = ρ ( \pi R1² × l ) / ∈o

so

E =  ρ ( R1²) / 2 ∈o R

6 0
4 years ago
Two beams of coherent light are shining on the same sheet of white paper. when referring to the crests and troughs of such waves
tensa zangetsu [6.8K]
Middle of the paper
5 0
4 years ago
A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm
ANEK [815]

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish  = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.

(B). the depth of the fish in the mirror.

d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm

Now find the depth of reflection of the fish in the bottom of the tank.

d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8  = 24.66\\

4 0
3 years ago
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