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3 years ago

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(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)

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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A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil

user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0

3 years ago

The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where

Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire = $R_T =\frac{\rho_T \ l}{A}$

Where $R_T$ =Resistance of wire at Temperature T

$\rho_T$ = Resistivity at temperature T $=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]$

Where $T_0 =20\ Deg\ C , \ \rho_0 = Constant, \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)$

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

$\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}$

$\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}$

$1.24=1+\alpha (T-20)$

$0.24=\alpha(\ T -20 )$

$Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1$

T = 81.52 Deg C

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3 years ago

A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 tu

Hoochie [10]

Answer:

The mutual inductance is $M = 0.000406 \ H$

Explanation:

From the question we are told that

The number of turns per unit length is $N = 1800$

The radius is $r = 0.0165 \ m$

The number of turns of the solenoid is $N_s = 210 \ turns$

Generally the mutual inductance of the system is mathematically represented as

$M = \mu_o * N * N_s * A$

Where A is the cross-sectional area of the system which is mathematically represented as

$A = \pi * r^2$

substituting values

$A = 3.142 * (0.0165)^2$

$A = 0.0008554 \ m^2$

also $\mu_o$ is the permeability of free space with the value $\mu_o = 4\pi * 10^{-7} N/A^2$

So

$M = 4\pi * 10^{-7} *1800 * 210 * 0.0008554$

$M = 0.000406 \ H$

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3 years ago

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3241004551 [841]

Answer:

yes

Explanation:

I think that air is made of particales because air is made out of gas but it has some particales in the air called aerosols.

7 0

3 years ago

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

Musya8 [376]

Answer: 3.12 * 10^12 F ( 3.12 pF)

Explanation: To calculate this capacitor of two hollow, coaxial, iron cylinders, we have to determine the potental differente between them and afeter that to use C=Q/ΔV

The electric field in th eregion rinner<r<router

By using the Gaussian law

∫E*ds=Q inside/εo

E*2*π*rinner^2*L= Q /εo

E=Q/(2*π*εo*r^2)

[Vab]=\int\limits^a_b {E} \, dr

where a and b are the inner and outer radii.

Then we have:

ΔV= 2*k*(Q/L)* ln (b/a)

replacing the values and using that C=Q/ΔV

we have:

C= L/(2*k*ln(b/a)=0.17/(2*9*10^9*3.023)=3.12 pF

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3 years ago