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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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Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that
kaheart [24]

Answer:

Young's modulus for the rope material is 20.8 MPa.    

 

Explanation:

The Young's modulus is given by:

E = \frac{FL_{0}}{A\Delta L}

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

F = T_{w} = W_{o}

Where:

T_{w}: is the tension of the wire

W_{o}: is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

F = mg = 1700 kg*9.81 m/s^{2} = 16677 N

Hence, the Young's modulus is:

E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa          

Therefore, Young's modulus for the rope material is 20.8 MPa.                

I hope it helps you!                                    

7 0
3 years ago
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

4 0
4 years ago
A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass
LiRa [457]
As density = mass/volume

So

Mass = density *volume

Mass = 25,365.4 * 2.7 = 68,486.58 g

<span>Mass of the granite monument to the nearest tenth = 68,485.6 g</span>

5 0
3 years ago
(True or False Statement)
sertanlavr [38]
False is correct......
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A motorcycle drives up a steeply inclined ramp. The work done on the motorcycle by the Earth’s gravitational force is ____. 1.
Leto [7]

Answer:

be

Explanation:

8 0
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