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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 plane wall of thickness 0.1 mm and thermal conductivity 25 W/m K having uniform volumetric heat generation of 0.3 MW/m3 is ins
juin [17]

Answer:

The maximum temperature is 90.06° C

Explanation:

Given that

t= 0.1 mm

Heat generation

q_g=0.3\ MW/m^3

Heat transfer coefficient

h=500\ W/m^2K

Here one side(left side) of the wall is insulated so the all heat will goes in to right side .

The maximum temperature will at the left side.

Lets take maximum temperature is T

Total heat flux ,q

q=q_g\times t

q=0.3\times 1000000\times 0.1 \times 10^{-3}\ W/m^2

q=30\ W/m^2

So the total thermal resistance per unit area

R=\dfrac{t}{K}+\dfrac{1}{h}

R=\dfrac{0.1\times 10^{-3}}{25}+\dfrac{1}{500}

R=0.002 K/W

We know that

q=ΔT/R

30=(T-90)/0.002

T=90.06° C

The maximum temperature is 90.06° C

3 0
3 years ago
An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over
Lady bird [3.3K]

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

5 0
3 years ago
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Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?
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Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

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The half-life of iodine-131 is 8.04 days. It’s decay constant equals
Svetllana [295]

Answer:

Explanation:

we know that half life of an element is

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λ=0.693/T

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λ=0.086

7 0
3 years ago
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