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

10. How much work is done by gravity if it pulls a 100 N object down 2 m?

NeTakaya

Answer:

200 J

Explanation:

W = f * d

W = 100N * 2m

W = 200 J

5 0

3 years ago

¿Cómo obtener razonadamente la primera ley de newton a partir de la segunda?

storchak [24]

Do you speak English? If so I can help you in the comments

5 0

3 years ago

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo

makkiz [27]

Answer:

(a) The speed of the target proton after the collision is: $V_{2f} =433(m/s)$ , and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$ .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle: $m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$ , and y axle: $0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$ . Now replacing the value given as: $v_{1i}=500(m/s)$ , $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$ , and assuming that $m_{1}=m_{2}$ , we get for x axle: $500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$ , then solving for $v_{2f}$ , we get: $v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get: $500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$ , now solving for $v_{1f}$ , we can find the speed of the projectile proton after the collision as: $v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$ , that is the speed of the target proton after the collision.

5 0

3 years ago

Which BEST describes the difference between speed and velocity?

kirill [66]

Answer : The correct option is, (D) Velocity includes rate of change and direction.

Explanation :

Speed : Speed is defined as the distance traveled by an object with respect to the time taken. It is a scalar quantity that means it tell us about the magnitude of an object not direction.

Velocity : Velocity is defined as the rate of change of position of an object with respect to the time. It is a vector quantity that means it tell us about the magnitude and direction of an object.

The only difference between the speed and the velocity is that the velocity tell us about magnitude and direction but speed tell us about magnitude only.

Hence, the correct option is, (D) Velocity includes rate of change and direction.

6 0

3 years ago

Read 2 more answers

Unas niñas en el receso estaban jugando a derramar flatulencias en un frasco de forma cilíndrica, cuyo radio tenía 5" y una altu

hram777 [196]

Answer:

Los gases siempre se adaptaran al volumen del contenedor en el que estén.

Entonces, si baja la presión del gas, la temperatura se mantendrá constante, lo que implica que tiene que aumentar la temperatura, pues para un gas ideal tenemos:

P*V = n*k*T

donde si el frasco esta cerrado, tenemos que n, k y V son constantes, entonces:

P = (n*k/V)*T

P = constante*T

(Notar que si el frasco no estuviera cerrado, entonces el numero n variaría de forma incontrolable, y este problema no se podría resolver de forma sencilla)

Entonces, si la presión baja, también baja la temperatura, pero el volumen se mantendrá constante, eso es lo importante.

Como el volumen se mantiene constante, el volumen que tomara el gas va a ser igual al volumen del frasco, sabemos que el volumen de un cilindro es:

V = (pi*r^2*h)

donde:

r = radio

pi = 3.14

h = altura.

en este caso, r = 5'' = 5 in

h = 7 in

Entonces el volumen sera:

V = 3.14*(5in)^2*7in = 549.5 in^2

3 0

3 years ago