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grin007 [14]
3 years ago
8

(Serway 9th ed., 6-27) The mass of a sports car is 1200 kg. The shape of the body is such that the aerodynamic drag coefficient

is 0.250 and the frontal area is 2.20 m2. Ignoring all other sources of friction, calculate the initial acceleration the car has if it has been traveling at 100 km/hr and is now shifted into neutral and allowed to coast. (Ans. 0.212 m/s2, opposite the velocity vector)
Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

a = - 0.248 m/s²

Explanation:

Frictional drag force

F = ½ *(ρ* v² * A * α)

ρ = density of air  , ρ = 1.295 kg/m^3

α = drag coef , α = 0.250

v = 100 km/h x 1000m / 3600s

v =  27.77 m/s

A = 2.20m^2

So replacing numeric in the initial equation

F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)

F = 298.6 N

Now knowing the force can find the acceleration

a = - F / m

a = - 298.6 N / 1200 kg

a = - 0.248 m/s²

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Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m

7 0
3 years ago
The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo
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Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

Distance=Speed\times Time

Thus distance covered in blinking of eye =

Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters

Thus no of football fields=\frac{936}{91.4}=9.15Fields

7 0
3 years ago
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

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We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various
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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}

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6 0
3 years ago
For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?
V125BC [204]

Answer:

The energy of photon, E=8\times 10^{-15}\ J

Explanation:

It is given that,

Voltage of anode, V=50\ kV=50\times 10^3\ V=5\times 10^4\ V

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

E=eV

e is charge of electron

E=1.6\times 10^{-19}\times 5\times 10^4

E=8\times 10^{-15}\ J

So, the maximum energy of the x- ray radiation is 8\times 10^{-15}\ J. Hence, this is the required solution.

6 0
3 years ago
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