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Shalnov [3]
2 years ago
14

Describe the effect of speed on voltage

Physics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0

Answer:

Speed has no effect on voltage.

Explanation:

The voltage of the battery in your car is always 12 to 13.5 volts. It makes no difference whether the car is home in the garage or zooming down the interstate.

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A stationary siren creates an 894 Hz
valentina_108 [34]

Answer:

12.3 m/s

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

Given:

fs = 894 Hz

fr = 926 Hz

c = 343 m/s

vs = 0 m/s

Find: vr

926 = 894 (343 + vr) / (343 + 0)

vr = 12.3

The speed of the car is 12.3 m/s.

5 0
2 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0
3 years ago
Read 2 more answers
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
2 years ago
A star’s parallax angle is 1.0. How far away is the star in light years?
larisa [96]
Distance = 2AU / tan1.0

If you mean 1.0 is in degrees, then Distance = 114.58 AU
7 0
2 years ago
Read 2 more answers
Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
Sergeu [11.5K]
The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
4 0
3 years ago
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