A.

A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured

nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0

3 years ago

En la Tierra, una balanza muestra que tu peso es 585 N.

ra1l [238]

Answer:

a) m = 59.63 [kg]

b) Wm = 95.41 [N]

Explanation:

El peso de un cuerpo se define como el producto de la masa por la aceleración gravitacional. DE esta manera tenemos:

W = m*g

Donde:

m = masa [kg]

g = gravedad = 9.81 [m/s^2]

m = W / g

m = 585 / 9.81

m = 59.63 [kg]

Es importante aclarar que la masa se conserva independientemente de la ubicación del cuerpo en el espacio.

Por ende su masa sera la misma en la luna.

El peso en la luna se calcula como Wm y es igual a:

Wm = 59.63 * 1.6 = 95.41 [N]

5 0

3 years ago

Which substances resist pH changes? <br> Acids<br> Hydrogen ions<br> Buffers<br> Bases

Vikki [24]

The correct choice is Buffers.

Buffers contains both acids and bases in equilibrium.when base is added to buffers, they adjust the pH value by releasing hydrogen ions.when acid is added to buffers, they adjust the pH value by consuming hydrogen ions. hence these substances resist any change in their pH value by releasing or absorbing hydrogen ions.This process keeps the pH value constant.

6 0

3 years ago

Read 2 more answers

If you weighted 130 lbs on Earth how much would you weigh on the moon?

In-s [12.5K]

Answer:

21

Explanation:

Weight on the moon is 16.5 % of weight on earth

Weight on moon = 0.165 * 130

Weight on moon = 21 lbs

3 0

3 years ago

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago