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Nikolay [14]
3 years ago
10

Help ASAP!!!!

Physics
1 answer:
goldfiish [28.3K]3 years ago
7 0

Answer:

My guess is that it would be B) basic

(if its wrong I really am sorry)

Have a nice evening :)

Explanation:

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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

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Read 2 more answers
Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,
dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

Lo que da;

\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

Lo que da;

\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

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The origin of the universe remains a question. true false
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True ..............................
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