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JulijaS [17]
3 years ago
12

A car accelerates uniformly in a straight line

Physics
1 answer:
erik [133]3 years ago
3 0

Answer:

21.59 m/s

Explanation:

recall that one of the equations of motions can be expressed as

v² = u² + 2as

where,

v = final velocity (we are asked to find this)

u = initial velocity = 0m/s (because it says that it starts from rest)

a = acceleration = 3.7m/s²

s = distance travelled = 63 m

simply substitute the known values above into the equation:

v² = u² + 2as

v² = 0² + 2(3.7)(63)

v² = 466.2

v = √466.2

v = 21.59 m/s

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The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.

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Newton's work on forces regarding motion can never be neglected by scientists. Sir Isaac Newton when he was alive, among several of his works he proposed the three laws guiding the forces of motion. In this question we are only going to be treating only one out of the three Newton's Law of motion and that is the second Law Of Newton's laws of motion.

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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

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.

6 0
3 years ago
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