Can someone please help me with science.

nlexa [21]

Answer: D

Explanation: Look at the figure, we can conclude that the correct answer is

X: Low potential energy

Y: High potential energy

Z: Flow of electrons

Because electrons flow where there is difference in potential energy. And electrons move from a region of high potential energy to a region of low potential energy.

Since the arrow is pointing to X, that means

Z is the flow of electrons, X is of low potential energy and Y is of high potential energy.

6 0

3 years ago

A mixture of 1.00g h2 and 1.00g he is placed in a 1.00l container at 27 celcius calculate the partial pressure of each gas and t

r-ruslan [8.4K]

PV = nRT moles of H2 = 1/2 = 0.5 moles of He = 1/4 =0.25 T = 273 + 27 partial pressure of H2 Px1 = 0.5x0.083x300 P=12.45 atmospheres PP of He px1 = 0.25x0.083x300 P =6 22 atmospheres Totla pressure = 6.22 + 12.45 = 18.68 atm

4 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

The sound Intensity level with intensity 10x is 10dB.

B) $I=100I_{0}$

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

$\beta=10log(100)$

β = 20

With intensity 100x, level is 20dB.

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

2 years ago

A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on

muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

$x=ucos\alpha t-(1/2)gt^{2}$ but the acceleration in the horizontal path is zero, hence we have

$x=ucos\alpha t$

Since the horizontal distance covered is 155m at 7.6secs, we have $ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1$

Also from the vertical path, the distance covered is expressed as

$y=usin\alpha t-(1/2)gt^{2}$

since the horizontal distance covered in 7.6secs is 195m, then we have

$y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2$

Hence if we divide both equation 1 and 2 we arrive at

$\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\$

Hence if we substitute the angle into the equation 1 we have

$ucos72.02=20.38\\u=66.02m/s$

Hence the initial velocity is 66.02m/s

3 0

3 years ago

How much time elapses from when the first bit starts to be created until the conversion back to analog begins

zimovet [89]

The question is not complete. Kindly find the complete question below:

Host A converts analog to digital at a = 58 Kbps Link transmission rate R = 1.9 Mbps Host A groups data into packets of length L = 112 bytes Distance to travel d = 931.9 km Propagation speed s = 2.5 x 108 m/s Host A sends each packet to Host B as soon as it gathers a whole packet. Host B converts back from digital to analog as soon as it receives a whole packet. How much time elapses from when the first bit starts to be created until the conversion back to analog begins? Give answer in milliseconds (ms) to two decimal places, normal rounding, without units (e.g. 1.5623 ms would be entered as "1.56" without the quotes)

Answer / Explanation

The answer is 19.65

6 0

3 years ago