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ludmilkaskok [199]

2 years ago

10

An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (aver

age).What should the plane's course shift be so that it will fly due south?

1 answer:

Igoryamba2 years ago

5 0

Answer:

It should fly 8° to west of south at 430km/h

Explanation:

According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.

$V_{w}*cos(45) = V_{A}*sin(\alpha )$ Solving for α:

α = 8.03°

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

From the law of logarithm we have that

$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

$= 14.15 + 82$

$Z_{26}= 96.15 dB$

4 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

Learn more about current and voltage:

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#LearnwithBrainly

4 0

3 years ago

Most interstellar clouds are: Most interstellar clouds are: much bigger than our solar system. similar in size to clouds in Eart

cupoosta [38]

Answer:

Most interstellar clouds are much bigger than our solar system.

Explanation:

An interstellar cloud refers:

It is generally an accumulation of gas, plasma, and dust in our and other galaxies.
It is basically a denser-than-average region of the interstellar medium (ISM).

Interstellar clouds can be large up to 106 solar masses

It is also often said to be the most massive entities in the galaxy.

Hence

we can say about Interstellar clouds,

They are much bigger than our solar system.

learn more about interstellar clouds here:

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4 0

1 year ago

Can someone please help me?

Alina [70]

Answer:

D would be it

Explanation:

cause none of the rest Makes sense to the book of the picture and I'm 100% sure =)

6 0

3 years ago

PLSSS HELP MEEEE

vekshin1

Answer:

Cu2+

Explanation:

7 0

3 years ago

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