All categories

3 years ago

A banked circular highway is designed for traffic moving

Physics

1 answer:

Nuetrik [128]3 years ago

3 0

Answer:

0.063

Explanation:

velocity of the car, v = 40 km/h = 11.11 m/s

radius, r = 200 m

Let the coefficient of friction is μ.

The coefficient of friction relates to the velocity on banked road is given by

$\mu =\frac{v^{2}}{rg}$

where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.

By substituting the values, we get

$\mu =\frac{11.11^{2}}{200\times 9.8}$

μ = 0.063

You might be interested in

You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

6 0

3 years ago

What type of friction prevents a pile of rocks from falling apart?

BlackZzzverrR [31]

A because it some type of friction

6 0

3 years ago

A transverse wave is traveling from right to left. what direction does the medium vibrate?

lisabon 2012 [21]

<span>In transverse waves, particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.</span>

5 0

3 years ago

Read 2 more answers

HELP!!! ASP EASY QUESTION! WILL GIVE BRAINEST IF U ANSWER!

Harlamova29_29 [7]

I believe the answer is Anther.
The Anther produces the pollen

5 0

3 years ago

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and 4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and 1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and 2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and 2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and 8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge Right Charge Resulting force(N)

1μC 4μC 90 N

4μC 1μC 90 N

2μC 2μC 90 N

1μC 2μC 45 N

1μC 8μC 180 N

2μC 8μC 360 N

5 0

3 years ago