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4 years ago

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Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the univers

e. The universe is very big and very old. The universe consists of matter.

1 answer:

cluponka [151]4 years ago

4 0

Answer:

Explanation:

That Universe Consists of Matter

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What are the two main<br> categories of particles and<br> antiparticles?

Mice21 [21]

Answer:

Particles can be classified as hadrons – baryons and mesons – and leptons, each with its anti-particle, and they should know that interactions between these particles can be described in terms of transfer of other particles known as vector bosons.

Explanation:

Your Welcome, if you could give me Brainlist I would appreciate it!

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3 years ago

Explain how DNA sequences can provide evidence of evolution.

NNADVOKAT [17]

If you were able to unstrand each piece of DNA, you would see differences in time which would be labeled as "evolution"

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3 years ago

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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

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force $N=W-F$ . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

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3 years ago

What's on the other side of a black hole

bearhunter [10]

Not sure, nobody knows hopefully soon we will know

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4 years ago

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