Ask question

Ask question

All categories

Natasha_Volkova [10]

3 years ago

11

A 1840-kg car travels on a banked, horizontal curve of diameter 225 m. Find the maximum safe speed if the coefficient of frictio

n between the tires and the road is 0.90 and the banking angle is 6.0°. Additionally, what net force does the car experience in this case?

1 answer:

BigorU [14]3 years ago

8 0

Answer:

35 m/s

Explanation:

m = 1840 kg

2R = 225 m

R = 112.5 m

μ = 0.9

θ = 6 degree

use the formula for the banking of road

$v = \sqrt{\frac{Rg(Sin\theta +\mu Cos\theta )}{Cos\theta -\mu Sin\theta }}$

$v = \sqrt{\frac{112.5\times 9.8(Sin6 +0.9 Cos6 )}{Cos6 -0.9Sin6 }}$

v = 35 m/s

You might be interested in

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$

8 0

3 years ago

You pull two blocks connected by a lightweight string across a horizontal table. Block 1 has a mass of 1.00 kg, and block 2 has

coldgirl [10]

Use google
-Just sayi’n don’t get mad

7 0

3 years ago

A truck going 15 kmôŹ°€h has a head-on collision with a small car going 30 kmôŹ°€h. Which statement best describes the situation

zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

5 0

3 years ago

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Citrus2011 [14]

Answer:

a

The height is $H = 6.74 \ m$

b

The horizontal distance is $D = 23.74 \ m$

Explanation:

From the question we are told that

The speed is $v = 15 \ m/s$

The angle is $\theta = 40^o$

The height of the cannon from the ground is h = 2 m

The distance of the net from the ground is k = 1 m

Generally the maximum height she reaches is mathematically represented as

$H = \frac{v^2 sin^2 \theta }{2 * g } + h$

=> $H = \frac{(15)^2 [sin (40)]^2 }{2 * 9.8} + 2$

=> $H = 6.74 \ m$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here s is the displacement which is mathematically represented as

s = [-(h-k)]

=> s = -(2-1)

=> s = -1 m

There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

a = -g = -9.8

$u = v sin (\theta)$

So

$-1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2$

=> $-4.9t^2 + 9.6418t + 1 = 0$

using quadratic formula to solve the equation we have

$t = 2.07 \ s$

Generally distance covered along the horizontal is

$D = v cos (40) * 2.07$

=> $D = 15 cos (40) * 2.07$

=> $D = 23.74 \ m$

7 0

3 years ago

A 43 kg box is being pushed and pulled by

m_a_m_a [10]

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

$155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]$

The negative sign means that the box accelerates in a negative direction to the left.

4 0

3 years ago