Physics Problems for 9 – 9 – 16

Base your answers to questions 1 and 2 on the information below.

In a drill during basketball practice, a player runs the length of the 30.-meter court and back. The player does this three times in 60. seconds.

Problem 1

The magnitude of the player’s total displacement after running the drill is
(1) 0.0 m
(2) 30. m
(3) 60. m
(4) 180 m

Base sus respuestas a las preguntas 1 y 2 en la siguiente información.

En una ejercicio durante una práctica de básquet, un jugador corre el largo ida y vuelta de una cancha de 30 metros. El jugador hace esto tres veces en 60 segundos.

La magnitud del desplazamiento total del jugador tras hacer el ejercicio es
(1) 0.0 m
(2) 30. m
(3) 60. m
(4) 180 m

Discussion and Solution:
In order to answer this question correctly, remember the definition of displacement.
Displacement is the distance from the starting point to the ending point.
Since the player started and stopped at the same point, the total displacement was 0.0 m.
Choice (1).

Problem 2

The average speed of the player during the drill is
(1) 0.0 m/s
(2) 0.50 m/s
(3) 3.0 m/s
(4) 30. m/s

La rapidez promedio del jugador durante el ejercicio es
(1) 0.0 m/s
(2) 0.50 m/s
(3) 3.0 m/s
(4) 30. m/s

Discussion and Solution:
Since the player went up and back three times, (s)he traveled 180 m (30 x 2 x 3).
The time took was 60. Seconds.
Dividing these, remembering that speed is calculated by dividing distance by time, we get 3.0 m/s (180 m / 60. Sec).
Choice (3).

Problem 3

A baseball is thrown at an angle of 40.0° above the horizontal. The horizontal component of the baseball’s initial velocity is 12.0 meters per second. What is the magnitude of the ball’s initial velocity?
(1) 7.71 m/s
(2) 9.20 m/s
(3) 15.7 m/s
(4) 18.7 m/s

Una bola de béisbol es lanzada a un ángulo de 40.0° por encima de la horizontal. La componente horizontal de la velocidad inicial de la bola de béisbol es 12.0 metros sobre segundo. ¿Cuál es la magnitud de la velocidad inicial de la bola?
(1) 7.71 m/s
(2) 9.20 m/s
(3) 15.7 m/s
(4) 18.7 m/s

Discussion and Solution:
Since we know the angle (40.0o) and the horizontal component (12.0 m/s), we should use the trig formula for cosine (cosine = adjacent / hypotenuse).
cosin 40.0o = (12.0 m/s) / H.
Solving for the hypotenuse (H), we get 15.7 m/s or choice (3).

Problem 4

A particle could have a charge of
(1) 0.8 x 10−19 C
(2) 1.2 x 10−19 C
(3) 3.2 x 10−19 C
(4) 4.1 x 10−19 C

Una particula puede tener una carga de
(1) 0.8 x 10-19 C
(2) 1.2 x 10−19 C
(3) 3.2 x 10−19 C
(4) 4.1 x 10−19 C

Discussion and Solution:
Remember, one elementary charge is 1.6 x 10-19 C.
Particles must either have that charge, or whole-number multiples thereof.
Choice (3).

Problem 5

Which object has the greatest inertia?
(1) a 15-kg mass traveling at 5.0 m/s
(2) a 10.-kg mass traveling at 10. m/s
(3) a 10.-kg mass traveling at 5.0 m/s
(4) a 5.0-kg mass traveling at 15 m/s

¿Cuál objeto tiene la mayor inercia?
(1) una masa de 15-kg viajando a 5.0 m/s
(2) una masa de 10-kg viajando a 10. m/s
(3) una masa de 10-kg viajando a 5.0 m/s
(4) una masa de 5.0-kg viajando a 15 m/s

Discussion and Solution:
Inertia is a measure of an object’s mass.
The greater the mass, the greater its inertia.
Choice (1).

 

Leave a Reply

Your email address will not be published. Required fields are marked *