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inessss [21]
3 years ago
15

The drag coefficient of a car at the design conditions of 1 atm, 25°c, and 90 km/h is to be determined experimentally in a large

wind tunnel in a full-scale test. the height and width of the car are 1.25 m and 1.65 m, respectively. if the horizontal force acting on the car is measured to be 230 n, determine the total drag coefficient of this car. take the density of air at 1 atm and 25°c as ρ = 1.164 kg/m3. (round the final answer to two decimal places.)
Physics
1 answer:
Anvisha [2.4K]3 years ago
8 0
Velocity = 90 km/h = 25 m/s

Drag coefficient, Cd = 2D/Density of air*Frontal area*sqr. velocity

Substituting,

Cd = (2*230)/(1.164*1.25*1.65*25^2) = 0.3066
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The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various
Schach [20]

Answer:

q  =  -461532.5 \ C

Explanation:

From the question we are told that

     The  electric filed is  E  =  102 \ N/C  

Generally according to Gauss law

=>   E  A  =  \frac{q}{\epsilon_o }

Given that  the electric field is pointing downward  , the equation become

    - E  A  =  \frac{q}{\epsilon_o }

Here   q is the excess charge on the surface of the earth

          A is the surface  area of the of the earth which is mathematically represented as

     A  =  4\pi r^2

Where r is the radius of the earth which has a value r = 6.3781*10^6 m

 substituting values

    A  = 4 * 3.142  *   (6.3781*10^6 \ m)^2

    A  =5.1128 *10^{14} \ m^2

So

   q  =  -E  * A  *  \epsilon _o

Here \epsilon_o s the permitivity of free space with value

          \epsilon_o  =  8.85*10^{-12} \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

     q  =  -102  * 5.1128 *10^{14}  *  8.85 *10^{-12}

     q  =  -461532.5 \ C

6 0
3 years ago
What cloud extending to the gravitational limits of the solar system would comets come from?.
ser-zykov [4K]

The Oort cloud extends to the gravitational limits of the solar system would comets come from.

<h3>What is oort cloud?</h3>

It is a hypothetical idea of a cloud of mostly frozen planetesimals that would orbit the Sun at distances between 2,000 and 200,000 AU.

The Oort cloud reaches the solar system's outer gravitational boundaries, where comets originate.

Hence cloud extending to the gravitational limits of the solar system will be oorto cloud.

To learn more about the oort cloud refer;

brainly.com/question/23368033

#SPJ1

4 0
2 years ago
An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad
Anna11 [10]

Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0
3 years ago
Why is it important to understand forces?
yanalaym [24]
If people never learned forces, there would be a major gap in the world and how it works, let alone in physics...
as much as you don't wanna admit it, force is everywhere and you see it if not use it EVERY day in your life, something as simple as driving a car down the street or too school, your using force of your wheels to move your car, which is moving you
3 0
3 years ago
The maximum wavelength an electromagnetic wave can have and still eject an electron from a copper surface is 264 nm .What is the
Tamiku [17]

Answer:

4.71 eV

Explanation:

For an electromagnetic wave with wavelength

\lambda=264 nm = 2.64\cdot 10^{-7} m

the energy of the photons in the wave is given by

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.64\cdot 10^{-7}m}=7.53\cdot 10^{-19} J

where h is the Planck constant and c the speed of light. Therefore, this is the minimum energy that a photon should have in order to extract a photoelectron from the copper surface.

The work function of a metal is the minimum energy required by the incident light in order to extract photoelectrons from the metal's surface. Therefore, the work function corresponds to the energy we found previously. By converting it into electronvolts, we find:

E=\frac{7.53\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=4.71 eV

3 0
3 years ago
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