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Olegator [25]
2 years ago
15

That minimum coefficient of static friction between the tyres and the road that will allow the car that will round safely.

Physics
1 answer:
liraira [26]2 years ago
3 0

Answer:

The answer is "0.13562748"

Explanation:

Please find the complete question in the attached file.

Using formula:

\bold{U_s=\frac{v^2}{rg}}

Given:

r=1.60 \times 10^2 \ m\\\\g=9.8\\\\v= 52.5\ \frac{km}{hr}= 14.583\ \frac{m}{s}

put the value into the above-given formula:

U_s= \frac{14.583^2}{160 \times 9.8}

    = \frac{212.663889}{1568}\\\\=0.13562748

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Define the difference between the rigid body problem and a single particle problem.
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3 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

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3 years ago
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A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i
dlinn [17]

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x  10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ =  9 x 10⁹ x  3 x 10⁻⁹ / r

r =3.34 x  10⁷ m

Thus, the distance between the charges will be 3.34 x  10⁷ m.

Learn more about  electric potential energy.

brainly.com/question/12645463

#SPJ1

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Which of the following is an electromagnetic wave?
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Your answer will be Radio Waves . 

That seems to be the only to make sense. Hope that helps u 
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