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agasfer [191]
2 years ago
13

a 1250 kg car traveling at a speed of 25.0 m/s rounds a 175 m radius curve. assuming the road is level, determine the coefficien

t of static friction between the car’s tires and the road.
Physics
1 answer:
Fiesta28 [93]2 years ago
4 0

Coefficient of static friction between the car’s tires and the road is 0.3

What is static friction ?

A force that holds an object at rest is called static friction. The definition of static friction is: The resistance people feel when they attempt to move a stationary object across a surface without actually causing any relative motion between their body and the surface they are moving the object across.

Give an illustration of static friction.

When something is sitting on a surface, static friction affects it. For instance, every time you place your foot on the trail when trekking in the woods, there is static friction between your shoes and the surface.

Assume that road is level

Friction force provides centripetal force to the car

The weight of the car is = 1250 kg

speed of the car is = 25 m/s

curve radius =r= 175m

we know the formula of centripetal force is given by

f_{s} =\frac{mv^{2} }{r}         where m = mass of the object

                                   v=  velocity

                                    r=  curve radius

and we know  the static force is given by f= μN

                                                                     here, N= mg

so now,

μmg=\frac{mv^{2} }{r}\\

now put the all give values then we get,

$$\mu_r=\frac{v^2}{r g}=\frac{(25 \mathrm{~m} / \mathrm{sec})^2}{(175 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{sec}^2\right)}$$

μ=\frac{625}{1715}

μ= 0.3 , This is the coefficient of friction between the car tire and the rode

To learn more about static friction, Visit:  brainly.com/question/13754413

#SPJ4

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A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0
2 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
A bowling ball is rolling in a straight line toward the pins at a speed of 2.2 m/s. Its momentum is 4.75 kg*m/s. What is the
Thepotemich [5.8K]

Answer: m= 2.16 kg

Explanation: Momentum is expressed in the following formula:

p = mv

Derive to find m:

m = p / v

= 4.75 kg.m/s / 2.2 m/s

= 2.16 kg

Cancel out m/s and the remaining unit is in kg.

3 0
3 years ago
Read 2 more answers
3 questions I WILL MARK BRAINLYEST 100 POINTS!!!!!!
sammy [17]

Answer:

19 on the first one

Explanation:

5 0
2 years ago
Is V directly proportional to I (From V=IR)??<br>Or is V inversely proportional to I (From P=IV) ???
ziro4ka [17]

First of all, I is proportional V according to the Ohm's Law. R is merely a constant you need to obtain an equation. However, it is true that R changes with temperature and pressure, therefore Ohm's Law is only applicable in an invariable environment. Also this constant R is different for different materials.

So, do not get confused.

Ohm's law is not a universal law, please remember that as well. Some materials do not follow it and we call them non-ohmic conductors. I hope I helped! ^-^

3 0
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