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Karolina [17]
3 years ago
7

Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia

prędkość chłopca podczas marszobiegu?
Physics
1 answer:
USPshnik [31]3 years ago
8 0

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

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Answer:

W = 46 J

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We need to find the angle between the two vectors Force vector and displacement vector.

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tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\

Then we find the angle β of the displacement vector

tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

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The absolute value of d will be:

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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

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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 °

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