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nalin [4]
3 years ago
7

Please write in complete sentences.

Physics
1 answer:
BabaBlast [244]3 years ago
8 0

Answer:

The density of a material affects the speed that a wave will be transmitted through it. In general, the denser the transparent material, the more slowly light travels through it.

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If the load is 13 cm from the fulcrum, how much effort is needed to lifth the load ? Help me.
snow_tiger [21]
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:

-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.

-- Then the effort needed to lift the load is

(the weight of the load) x (13 / the distance between the fulcrum and the effort)
7 0
3 years ago
Lithium (Li) has a charge of +1, and oxygen (O) has a charge of –2. Which is the chemical formula? LiO2 LiO Li2O Li2O2
LiRa [457]

Answer:

The results is                  Li₂O

Explanation:

n a chemical reaction, the electron miner in the formula must be balanced, that is, the number of positive and negative charges is the same

In this case, oxygen has a valence of -2 that is, it can give two electrons and lithium has a valence of +1, therefore two lithium atoms are needed for the charges to be neutral, therefore the chemical formula is

The results is                  Li₂O

6 0
3 years ago
Read 2 more answers
The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations d
ozzi
The solution for this problem is: In the figure, you now know that total length of the kerosene column
So at x – xPatm + Pkg(H0 th) = Pa + Pwgh
Now H0 + h = 20 + 91.1 mm = 111.1 mm
Therefore = Pkg 0.1111 – P2g= h = 56 x 0.111 – 98 / 1000 x 9.81= 0.081 m or 81 mn
Therefore H0 = 111.1 - 81= 30.1 mm
6 0
3 years ago
1 point
BaLLatris [955]

Answer:1

Explanation:

7 0
3 years ago
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the d
frez [133]

The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

\beta=10log(\frac{I}{I_{0}} )dB,

where I_{0} is a reference intensity. for sound waves, I_{0} is taken to be 10^{-12} W/m^{2}. Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. I=10I_{0}? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. I=100I_{0}? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels (\Delta \beta_{2},\Delta \beta_{4} and \Delta \beta_{8}) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) I=10I_{0}

\beta=10log(\frac{10I_{0}}{I_{0}} )

\beta=10log(10 )

β = 10

<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.

B) I=100I_{0}

\beta=10log(\frac{100I_{0}}{I_{0}} )

\beta=10log(100)

β = 20

<u>With intensity 100x, level is </u><u>20dB</u>.

C) To calculate the change, take the f to be the factor of increase:

For \Delta \beta_{2}:

I=2I_{0}

\beta=10log(\frac{2I_{0}}{I_{0}} )

\beta=10log(2)

β = 3

For \Delta \beta_{4}:

I=4I_{0}

\beta=10log(\frac{4I_{0}}{I_{0}} )

\beta=10log(4)

β = 6

For \Delta \beta_{8}:

I=8I_{0}

\beta=10log(\frac{8I_{0}}{I_{0}} )

β = 9

Change is

\Delta \beta_{2},\Delta \beta_{4}, \Delta \beta_{8} = 3,6,9 dB

6 0
3 years ago
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