(a) Equating centripetal force to friction force, one finds the relation
v² = kar
for car speed v, coefficient of friction k, radius of curvature r, and downward acceleration a.

There is already downward acceleration due to gravity. The additional accceleration due to the wing is
a = F/m = 10600 N/(805 kg) ≈ 13.1677 m/s²
We presume this is added to the 9.80 m/s² gravity provides, so the coefficient of friction is
k = v²/(ar) = (54 m/s)²/((13.1677 m/s² +9.80 m/s²)·(155 m))
k ≈ 0.8191

(b) The maximum speed is proportional to the square root of the downward acceleration. Changing that by a factor of 9.80/(9.80+13.17) changes the maximum speed by the square root of this factor.
max speed with no wing effect = (54 m/s)√(9.8/22.97) ≈ 35.27 m/s

8 0

3 years ago

The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per

Natasha_Volkova [10]

Answer:

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons

Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

$I = \frac{q}{t}$

$I = \frac{Ne}{t}$

$I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}$

$I = 0.96 \ A$

The magnitude of the electric field is:

E = $\frac{I}{ \mu n eA}$

E = $\frac{I}{ \mu n e(\pi r^2)}$

E = $\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}$

E = 0.2631 N/C

8 0

3 years ago

Physics!!! Please help!!!!!!!

Marianna [84]

Answer:

it is a very good morning amor de g the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of the first paragraph of 88th paragraph of the first paragraph

8 0

3 years ago