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Naddik [55]
3 years ago
14

As an emergency vehicle approaches Bob and moves away from Jill, how does the actual frequency of the siren change? A) As an eme

rgency vehicle approaches Bob the frequency increases. B) As an emergency vehicle moves away from Jill the frequency decreases. C) The actual frequency of the siren does not change despite appearances. D) Bob does not notice any change until the vehicle passes him and the siren increases.
Physics
2 answers:
asambeis [7]3 years ago
8 0

its C on USAtestprep

Tamiku [17]3 years ago
7 0
The correct answer is: 
<span>C) The actual frequency of the siren does not change despite appearances.

In fact, Bob will observe an increase in the apparent frequency as the emergency vehicle approaches him, while Jill will observe a decrease in the apparent frequency as the emergency vehicle moves away from him, because of the Doppler effect (the relative velocity between the observer and the source of the sound is changing), but this effect involves the apparent frequency, while the real frequency of the siren will remain the same.</span>
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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of
levacccp [35]

\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

\:  \:  \:

<u>Let's Slove :</u><u> </u>

  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

\:  \:

  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
3 0
1 year ago
(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass
11Alexandr11 [23.1K]

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants.  Mass of the Earth: 5.97 &times; 10^24  kg (assume a uniform mass distribution)  Radius of the Earth: 6371 km  Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?  

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

so  

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

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KonstantinChe [14]

Answer:

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Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

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Its B.hope this helps

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A soccer ball would keep moving forever without physics, because without force to act upon the soccer ball, it could, or will not be able to stop the acceleration. And force is a factor in physics.
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