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Effectus [21]
2 years ago
9

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

increases its speed at uniform rate of at ≡ d |v| dt = 5.22 m/s 2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 25.6 m/s, what is the coefficient of static friction between the tires and the road?
Physics
1 answer:
marysya [2.9K]2 years ago
4 0

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) =  mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

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n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
A type of Lead (Pb) ion has a + 4 oxidation number. Sulfur (S) has a — 2 oxidation number. What would be the chemical formula fo
serious [3.7K]

Answer: The chemical formula for the compound of these two elements is PbS_2

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal lead is having an oxidation state of +4 called as Pb^{4+} cation and sulphur non metal has oxidation state of -2 called as S^{2-}. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral PbS_2

The chemical formula for the compound of these two elements is PbS_2

4 0
3 years ago
A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o
defon

Part a)

Magnitude of electric field is given by force per unit charge

E = \frac{F}{q}

E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}

E = 2150 N/C

Part b)

Electrostatic force on the proton is given as

F = qE

F = 1.6 * 10^{-19} * 2150

F = 3.44 * 10^{-16} N

PART C)

Gravitational force is given by

F_g = mg

F_g = 1.6 * 10^{-27}*9.8

F_g = 1.57 * 10^{-26} N

PART d)

Ratio of electric force to weight

\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}

\frac{F_e}{F_g} = 2.2 * 10^{10}

7 0
3 years ago
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CaHeK987 [17]
The price of coast to coast membership in united states could lie anywhere between $2,000 to $ 5,000
Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price
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Which statement best explains why an object
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D, Mercury as a weaker gravitational pull! Due to mercury being farther from the sun and it being a smaller planet it has a weaker pull
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