All categories

3 years ago

The coefficients of friction between a race cars tyres and the track surface are

1 answer:

5 0

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

please mark me as a brainlieast

You might be interested in

If you ran 15 km/hr for 2.5 hours, how much distance would you cover?

SVETLANKA909090 [29]

Answer: 37.5 km

Explanation:

The question is that

If you ran 15 km/hr for 2.5 hours, how much distance would you cover ?

Where

Speed = 15 km/ hr

Time = 2.5 hours

Using the formula for speed.

Speed = distance/time

Substitute speed and time into the formula

15 = distance/ 2.5

Make distance the subject of formula by cross multiplying.

Distance = 15 × 2.5

Distance = 37.5 km.

6 0

3 years ago

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured

kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

Mass of the machine, m=70 kg
Stiffness of the system, k=30000 N/m
Damping ratio=0.2
Damping force, F=450 N
Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\$

Where

$\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

$\omega=13\ \rm rad/s$
$\rho=0.2$

$F_0=450\ \rm N$
$m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$

d)The magnitude of the force transmitted to the ground is

$F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N$

e)The maximum velocity is given by $V_{max}$

$V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s$

6 0

3 years ago

Inessa05 [86]

Savings account is where she can have access to her money anytime in case of an emergency.

<h3>What is a Savings account?</h3>

This is a type of account run by a financial institution which yields interests on deposits made.

This type of account usually comes with features in which the money can be withdrawn via an ATM card or by visiting the bank to collect it. This therefore makes it the most appropriate choice for Analisa.

Read more about Savings account here brainly.com/question/25787382

7 0

2 years ago

A 70-kg astronaut (including spacesuit and equipment) is floating at rest a distance of 13 m from the spaceship when she runs ou

Keith_Richards [23]

Answer:

Explanation:

mass of the astronaut including the spacesuit, $M=30$