Can someone please help me I don’t understand this

A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a

skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at $58 km/h\approx 16.11 m/s$

Radius of the curve 201 m

Actual traffic velocity = $37 km/h approx 10.27 m/s$

For banking of road

$tan\theta =\frac{v^2}{rg}$

$tan\theta =\frac{16.11^2}{201\times 9.81}$

$\theta =7.49^{\circ}$

Centripetal acceleration is given by

$a=\frac{v^2}{r}$

Taking component of centripetal acceleration

along and perpendicular to surface

$a_{parallel}=\frac{v^2cos\theta }{r}$

$a_{perpendicular}=\frac{v^2sin\theta }{r}$

From FBD

$mgsin\theta -f_s=ma_{parallel}$

$f_s=mgsin\theta -ma_{parallel}$ ----1

where $f_s$ is frictional force

$N-mgcos\theta =ma_{perpedicular}$

$N=mgcos\theta +ma_{perpedicular}$ ----2

and we know coefficient of friction is given by

$\mu =\frac{f_s}{N}$

$\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}$

$\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}$

$\mu =\frac{1.2804-0.5202}{9.726+0.068}$

$\mu =0.077$

7 0

3 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago

A hunter is aiming horizontally at a monkey who is sitting in a tree. The monkey is so terrified when it sees the gun that it fa

Dvinal [7]

Answer:a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.

Explanation: The bullet keeps as it aim( the monkey) unless it is redirected by an external force that could redirect it. Hence, the bullet speeds straight forward.

3 0

3 years ago

two teams are playing tug of war. team a pulls to the right with a force of 450n .team b pulls to the left with a force of 415 n

Alex_Xolod [135]

Explanation:

It is given that, two teams are playing tug of war.

Force applied by Team A, $F_A=450\ N$

Force applied by Team B, $F_B=415\ N$

We need to find the net force acting on the rope. It is equal to :

$F_{net}=F_A-F_B$

$F_{net}=450-415$

$F_{net}=35\ N$

So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.

4 0

3 years ago

If a vibrating string is made shorter (i.e., by holding your finger on it), what effect does

Ghella [55]

<span>Pitch and frequency are more or less the same thing - high pitch = high frequency.
The freqency of vibration of a string f = 1/length (L) so as length decreases the frequency increases.</span>

4 0

3 years ago