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lilavasa [31]
3 years ago
5

O D. Both objects won't move at all. They will just stay where they were released.

Physics
1 answer:
astraxan [27]3 years ago
6 0

Answer:

Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.

Probably D:  Soccer ball was affected by air resistance more than the golf ball.

Explanation:

Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.

In a vacuum, (no friction) both objects fall at the same rate regardless of mass.

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PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV b
masha68 [24]

Answer:

5694000 min

Explanation:

Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.

(4 h/day) × (60 min/1 h) = 240 min/day

They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.

240 min/day × (365 day/year) = 87600 min/year

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4 years ago
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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