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Vinil7 [7]
3 years ago
15

Explain how Law 1 applies to the image to the left.

Physics
2 answers:
Margarita [4]3 years ago
6 0

Answer:

12?

Explanation:

alisha [4.7K]3 years ago
3 0

Answer:

12

Explanation:

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Energy that tracked in waves though matter is?
kolbaska11 [484]

Answer:

sound eneegy pls

brainly

8 0
3 years ago
Elements are arranged in groups by similar atomic structure on the periodic table. This allows for an elements properties to be
DiKsa [7]

C

Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.

Explanation:

This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has  a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius

The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.

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5 0
4 years ago
Read 2 more answers
PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat
Svetach [21]
Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of  1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C
4 0
4 years ago
An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th
AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × 10^{-6} m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top  t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = 10^{5} Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

so

pressure at bottom P2 = 10^{5} + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × 10^{5}  Pa

so from gas law

\frac{P1*V1}{t1} = \frac{P2*V2}{t2}

here p is pressure and v is volume and t is temperature

so put here value and find v1

\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0
3 years ago
the fundamental frequency for the 3rd chord on a five-string guitar is 240 Hz. what frequency would produce the 10th harmonic?​
Mamont248 [21]

The frequency of the 10th harmonic is 800 Hz

Explanation:

The frequency of the nth-harmonic for the standing waves in a string is given by the equation

f_n = n f_1

where

f_1 is the fundamental frequency of the string

In this problem, we are given the frequency of the 3rd harmonic:

f_3 = 240 Hz

Which can be rewritten in terms of the fundamental frequency

f_3 = 3 f_1

So we find f_1:

f_1 = \frac{f_3}{3}=\frac{240}{3}=80 Hz

Now that we have the fundamental frequency, we can find the frequency of the 10th harmonic, with n = 10 :

f_{10} = 10f_1 = (10)(80)=800 Hz

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6 0
4 years ago
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