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3 years ago

6

A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular

arc of 230 m radius. The roadway is banked so that a vehicle can go around the curve with the friction force from the road equal to zero. At what angle is the roadway banked?

1 answer:

Fynjy0 [20]3 years ago

3 0

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

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Read 2 more answers

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV

Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

$= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

$= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

8 0

3 years ago

A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an

ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

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complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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2 years ago