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katovenus [111]
3 years ago
5

A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8

2, what is the greatest speed the car can have in the corner without skidding?
Physics
1 answer:
Vaselesa [24]3 years ago
5 0

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

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Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

where

C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

Therefore,

A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

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3 years ago
After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample
alina1380 [7]

Answer:

4 days

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n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

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Svetlanka [38]

Answer:

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Explanation:

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