Physics Problems for 9 – 30 – 16

Problem 1

A distance of 1.0 × 10−2 meter separates successive crests of a periodic wave produced in a shallow tank of water. If a crest passes a point in the tank every 4.0 × 10−1 second, what is the speed of this wave?
(1) 2.5 × 10−4 m/s
(2) 4.0 × 10−3 m/s
(3) 2.5 × 10−2 m/s
(4) 4.0 × 10−1 m/s

Una distancia de 1.0×102 metro separa crestas sucesivas de una onda periódica producida en un tanque de vacío de agua. Si una cresta pasa un punto en el tanque cada 4.0×101 segundo, ¿cuál es la rapidez de esta onda?
(1) 2.5 x 104 m/s
(2) 4.0 x 103 m/s
(3) 2.5 x 102 m/s
(4) 4.0 x 101 m/s

Discussion and Solution:
Velocity is calculated by dividing distance travelled by the time of travel:

v = d/t.

Substituting in what we know, we get:

v = (1.0 x 10-2 m)/(4.0 x 10-1 s) = 2.5 x 10-2 m/s.

Choice (3).

Problem 2

The electronvolt is a unit of
(1) energy
(2) charge
(3) electric field strength
(4) electric potential difference

El electronvoltio es una unidad de
(1) energía
(2) carga
(3) intensidad de campo eléctrico
(4) diferencia potencial eléctrica

Discussion and Solution:
Straight definition/memorization.
Choice (1).

Problem 3

Which particle would produce a magnetic field?
(1) a neutral particle moving in a straight line
(2) a neutral particle moving in a circle
(3) a stationary charged particle
(4) a moving charged particle

¿Cuál partícula produciría un campo magnético?
(1) una partícula neutral moviéndose en una línea recta
(2) una partícula neutral moviéndose en un círculo
(3) una partícula cargada estacionaria
(4) una partícula cargada en movimiento

Discussion and Solution:
When charged particles move (e.g., electrons in a wire), a magnetic field is produced. In the same manner, when a magnetic field is moved, an electric current is produced.
Choice (4).

Problem 4

A physics student takes her pulse and determines that her heart beats periodically 60 times in 60 seconds. The period of her heartbeat is
(1) 1 Hz
(2) 60 Hz
(3) 1 s
(4) 60 s

Una estudiante de física toma su pulso y determina que su corazón late periódicamente 60 veces en 60 segundos. El período de los latidos del corazón es
(1) 1 Hz
(2) 60 Hz
(3) 1 s
(4) 60 s

Discussion and Solution:
Period is defined as the time for one event to occur. To do this, we divide the total time by the total number of occurrences:

(60 s)/(60) = 1 s.

Choice (3).

Problem 5

Moving 4.0 coulombs of charge through a circuit requires 48 joules of electric energy. What is the potential difference across this circuit?
(1) 190 V
(2) 48 V
(3) 12 V
(4) 4.0 V

Mover 4.0 coulombs de carga a través de un circuito requiere 48 joules de energía eléctrica. ¿Cuál es la diferencia potencial a través de este circuito?
(1) 190 V
(2) 48 V
(3) 12 V
(4) 4.0 V

Discussion and Solution:
Use the formula, V = W/q, where V = potential difference, W = work, and q = charge.
Substituting in what we know, we get:

V = W/q = (48 J)/ (4.0 C) = 12 V

Choice (3).

 

 

Chemistry Problems for 9 – 28 – 16

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

The equation below represents an equilibrium system of SO2(g), O2(g), and SO3(g).

The reaction can be catalyzed by vanadium or platinum.

2SO2(g) + O2(g) ↔ 2SO3(g) + energy

Base sus respuestas a las preguntas de la 1 y 2 en la siguiente información y su conocimiento de química.

La siguiente ecuación representa un sistema en equilibrio compuesto de SO2(g), O2(g), y SO3(g). La reacción puede ser catalizada por vanadio o platino.

2SO2(g) + O2(g) ↔ 2SO3(g) + energía

Problem 1

Compare the rates of the forward and reverse reactions at equilibrium.

Compare las velocidades de las reacciones directa y reversa en equilibrio.

Discussion and Solution:

Since this is a reversible reaction, at equilibrium, the rates of the forward and reverse reactions are equal. The rates are the same.

Problem 2

State how the equilibrium shifts when SO3(g) is removed from the system.

Exponga como el equilibrio cambia cuando el SO3(g) es removido del sistema.

Discussion and Solution:

When SO3(g) is removed from the system: The equilibrium will shift to favor the formation of SO3; The rate of the forward reaction is greater than the rate of the reverse reaction; The equilibrium will shift to favor the forward reaction; The equilibrium will shift to the right; and The concentrations of the reactants will decrease.

Problem 3

Determine the volume of 2.00 M HCl(aq) solution required to completely neutralize 20.0 milliliters of 1.00 M NaOH(aq) solution.

Determine el volumen de una solución de 2.00 M HCl(aq) requerido para neutralizar completamente 20.0 mililitros de una solución de 1.00 M NaOH(aq).

Discussion and Solution:

We need to use the following equation, where Ma = molarity of the acid, Va = volume of the acid, Mb = molarity of the base, and Vb = volume of the base:

MaVa = MbVb

Since we are looking for Va, we divide both sides by Ma and substitute in what values we know and then simplify.

Va =  = 10.0 ml

Problem 4

Given the balanced equation representing a reaction:

6Li + N2 → 2Li3N

Which type of chemical reaction is represented by this equation?
(1) synthesis
(2) decomposition
(3) single replacement
(4) double replacement

Dada la ecuación balanceada representando una reacción:

6Li + N2 → 2Li3N

¿Qué tipo de reacción química es representada por esta ecuación?
(1) síntesis
(2) decomposición
(3) desplazamiento simple
(4) doble desplazamiento

Discussion and Solution:

Since we are starting off with two elements (Li and N) and ending up with a compound (Li3N), this is an example of a synthesis reaction.

Choice (1)

In choice (2), we would have to start off with a compound and end up with two elements.

In choice (3), we would have to start off with an element and a compound. The element would then replace one of the elements found in the compound.

In choice (4), we would start off with two compounds, who would switch elements with each other.

Problem 5

A rigid cylinder with a movable piston contains 50.0 liters of a gas at 30.0°C with a pressure of 1.00 atmosphere. What is the volume of the gas in the cylinder at STP?
(1) 5.49 L
(2) 45.0 L
(3) 55.5 L
(4) 455 L

Un cilindro rígido con un pistón movible contiene 50.0 litros de un gas a 30.0°C con una presión de un 1 atmósfera. ¿Cuál es el volumen del gas en el cilindro en STP?
(1) 5.49 L
(2) 45.0 L
(3) 55.5 L
(4) 455 L

Discussion and Solution:

For this we would use the combined gas law, where VI would be the initial volume of the gas, VF would be the final volume, PI would be the initial pressure, PF would be the final pressure, TI would be the initial temperature (in Kelvin), and TF would be the final temperature (in Kelvin).

VF =  = 45.0 L

Choice (2)

 

Physics Problems for 9 – 23 – 16

Base your answers to questions 1 through 4 on the information below and on your knowledge of physics.

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a resistor. The circuit does not contain a voltmeter or an ammeter. When the circuit is operating, the total current through the circuit is 0.50 ampere.

Problem 1

Base sus respuestas a las preguntas de la 1 a la 4 en la siguiente información y su conocimiento de física.

Un estudiante construyó un circuito en serie que consiste en una batería de 12.0 voltios, una lámpara de 10.0 ohm y una resistencia. El circuito no contiene un voltímetro ni un amperímetro. Cuando el circuito está operativo, la corriente total que pasa a través del circuito es 0.50 amperios.

Discussion and Solution:

 

Problem 2

Determine the equivalent resistance of the circuit.

Determine la resistencia equivalente del circuito.

Discussion and Solution:

V = IR

R = V / I = (12.0 V) / (0.5 A) = 24Ω

Problem 3

Determine the resistance of the resistor.

Determine la resistencia del resistor.

Discussion and Solution:

RT = RA + RL

RA = RTRL = 24 Ω – 10.0 Ω = 14Ω

Problem 4

Calculate the power consumed by the lamp.

Calcule el poder consumido por la lámpara.

Discussion and Solution:

P = I2R = (0.50 A)2(10.0 Ω) = 2.5W

or

P = VI = (5.0 V)(0.50 A) = 2.5W

Base your answers to question 5 on the information below and on your knowledge of physics.

Pluto orbits the Sun at an average distance of 5.91 × 1012 meters. Pluto’s diameter is 2.30 × 10meters and its mass is 1.31 × 1022 kilograms.

Charon orbits Pluto with their centers separated by a distance of 1.96 × 107 meters. Charon has a diameter of 1.21 × 106 meters and a mass of 1.55 × 1021 kilograms.

Problem 5

Calculate the magnitude of the gravitational force of attraction that Pluto exerts on Charon.

Base sus respuestas a las preguntas de la 5 en la siguiente información y su conocimiento de física.

Plutón orbita el Sol a un distancia promedio de 5.91 x 1012 metros. El diámetro de Plutón es 2.30 x 106 metros y su masa es de 1.31 x 1022 kilogramos.

Caronte orbita Plutón con sus centros separados por una distancia de 1.96 x 107 metros. Caronte tiene un diámetro de 1.21 x 106 metros y una masa de 1.55 x 1021 kilogramos.

Calcule la magnitud de la fuerza gravitacional de atracción que ejerce Plutón en Caronte.

Discussion and Solution:

Fg = Gm1m2 / r2

Fg = (6.67 x 10-11 N•m2/kg2)(1.31 x 1022 kg)(1.55 x 1021 kg) / (1.96 x 107 m)2

Fg = 3.53 x 1018 N

 

Chemistry Problems for 9 – 21 – 16

 

Base your answers to questions 1 through 3 on the information below and on your knowledge of chemistry.

When magnesium is ignited in air, the magnesium reacts with oxygen and nitrogen.

The reaction between magnesium and nitrogen is represented by the unbalanced equation below.

Base sus respuestas a las preguntas de la 1 a la 3 en la siguiente información y su conocimiento de química.

Cuando el magnesio es encendido en el aire, el magnesio reacciona con oxígeno y nitrógeno.

La reacción entre magnesio y nitrógeno está representada por la siguiente ecuación desbalanceada.

Mg(s) + N2(g) → Mg3N2(s)

Problem 1

Balance the equation for the reaction between magnesium and nitrogen, using the smallest whole-number coefficients.

Balancee la ecuación para la reacción entre el magnesio y el nitrógeno, usando el número entero de coeficientes más pequeño.

Discussion and Solution:

On the product side, there are three magnesium atoms, hence, we need three magnesium atoms on the reactant side.

When we insert a coefficient of three for the magnesium, the equation is balanced.

3Mg(s) + N2(g) → Mg3N2(s).

 

Problem 2

In the ground state, which noble gas has atoms with the same electron configuration as a magnesium ion?

En estado fundamental, ¿cuál gas noble tiene átomos con la misma configuración de electrón que un ion de magnesio?

Discussion and Solution:

In becoming an ion, a magnesium atom takes on the same electron configuration as Ne or neon.

Problem 3

Explain, in terms of electrons, why an atom of the metal in this reaction forms an ion that has a smaller radius than its atom.

Explique, en términos de electrones, porque un átomo del metal en esta reacción forma un ion que tiene un radio más pequeño que su átomo.

Discussion and Solution:

Acceptable responses would include any of the following, but are not limited to just them:

An atom of magnesium loses its outer shell electrons to form the Mg2+ ion.

The electron configuration of a magnesium atom is 2-8-2, and the electron configuration of the magnesium ion is 2-8.

An atom of the metal loses electrons to form the ion.

Base your answers to questions 4 through 5 on the information below and on your knowledge of chemistry.

The balanced equation below represents a reaction.

Base sus respuestas a las preguntas de la 4 a la 5 en la siguiente información y su conocimiento de química.

La siguiente ecuación balanceada representa una reacción.

O2(g) + energy → O(g) + O(g)

Problem 4

Identify the type of chemical bond in a molecule of the reactant.

Identifique el tipo de enlace químico en una molécula del reactante.

Discussion and Solution:

Acceptable responses would include any of the following, but are not limited to just them:

covalent

double covalent

nonpolar

double

Problem 5

Explain, in terms of bonds, why energy is absorbed during this reaction.

Explique, en términos de enlaces, porque la energía es absorbida durante esta reacción.

Discussion and Solution:

Energy is needed to break the bonds in O2.

 

Math Problems for 9 – 19 – 16

 

Problem 1

What is the product of 2x + 3 and 4x2 – 5x + 6?
(1) 8x3 – 2x2 + 3x + 18
(2) 8x3 – 2x2 – 3x + 18
(3) 8x3 + 2x2 – 3x + 18
(4) 8x3 + 2x2 + 3x + 18

¿Cuál es el producto de 2x + 3 y 4x2 – 5x + 6?
(1) 8x3 – 2x2 + 3x + 18
(2) 8x3 – 2x2 – 3x + 18
(3) 8x3 + 2x2 – 3x + 18
(4) 8x3 + 2x2 + 3x + 18

Discussion and Solution:
To solve this problem, first multiply the second expression by 2x. Then multiply the second expression by the 3. Finally, combine the like terms.

2x(4x2 – 5x + 6) = 8x3 – 10x2 + 10x

3(4x2 – 5x + 6) = 12x2 – 15x + 18

(8x3 – 10x2 + 10x) + (12x2 – 15x + 18)

8x3 + 2x2 – 3x + 18

Choice (3).

Problem 2

What are the solutions to the equation 3x2 + 10x = 8?
(1)  2/3 and – 4
(2) – 2/3 and 4
(3)  4/3 and – 2
(4) – 4/3 and 2

¿Cuáles son las soluciones para la ecuación 3x2 + 10x = 8?
(1)  2/3 and – 4
(2) – 2/3 and 4
(3)  4/3 and – 2
(4) – 4/3 and 2

Discussion and Solution:
To solve this equation, first subtract 8 from each side.

3x2 + 10x – 8 = 8 – 8
3x2 + 10x – 8 = 0

Since the only factors for the coefficient 3 are 3 and 1, the first components for the factors are:

(3x      )(x     )

Since there is a negative sign in front of the 8, we get:

(3x +    )(x –    ) or (3x –    )(x +    )

The only factors for the 8 are (8 and 1) and (4 and 2).

That gives us the following as possible solutions:

(3x + 8)(x – 1)

(3x – 8)(x + 1)

(3x + 4)(x – 2)

(3x + 2)(x – 4)

(3x – 4)(x + 2)

(3x – 2)(x + 4)

Using FOIL, the only one of the above expressions that would result in

3x2 + 10x – 8 is:

(3x – 2)(x + 4)

Setting each equal to 0 and solving, we get:

(3x – 2) = 0

3x = 2

x = 2/3

 

(x + 4) = 0

x = – 4

 

Choice (1).

Problem 3

A system of equations is given below.

x + 2y = 5

2x + y = 4

Which system of equations does not have the same solution?

(1) 3x + 6y = 15

2x + y = 4

(2) 4x + 8y = 20

2x + y = 4

(3) x + 2y = 5

6x + 3y = 12

(4) x + 2y = 5

4x + 2y = 12

A continuación se muestra un sistema de ecuaciones.

 x + 2y = 5

2x + y = 4

¿Qué sistema de ecuaciones no tiene la misma solución?

(1) 3x + 6y = 15

2x + y = 4

(2) 4x + 8y = 20

2x + y = 4

(3) x + 2y = 5

6x + 3y = 12

(4) x + 2y = 5

4x + 2y = 12

Discussion and Solution:
Solution set (1) involves the first equation multiplied by 3, hence they have the same solution.
Solution set (2) involves the first equation multiplied by 4, hence they have the same solution.
Solution set (3) involves the second equation multiplied by 3, hence they have the same solution.
Solution set (4) involves different equation, hence they have different solutions.

Choice (4).

Problem 4
Find the zeros of f(x) = (x – 3)2 – 49, algebraically.

Encuentre los ceros de f(x) = (x – 3)2 – 49, algebraicamente.

Discussion and Solution:
In order for the function to be zero, (x – 3)2 must equal 49.

(x – 3)2 = 49

x – 3 = 7

x = 10

(x – 3)2 = 49

x – 3 = – 7

x = – 4

Problem 5

Solve the equation below for x in terms of a.

4(ax + 3) – 3ax = 25 + 3a

 Resuelva la siguiente ecuación para x en términos de a.

4(ax + 3) – 3ax = 25 + 3a

Discussion and Solution:

First remove the parentheses, then combine like terms:

4(ax + 3) – 3ax = 25 + 3a

4ax + 12 – 3ax = 25 + 3a

ax + 12 = 25 + 3a

ax = 13 + 3a

Divide both sides by a:

ax = 13 + 3a

x = (13 + 3a) / a

or

x = 13/a  + 3

 

 

 

 

Physics Problem for 9 – 16 – 16

 

Problem 1

What is the speed of light (f = 5.09 × 1014 Hz) in ethyl alcohol?
(1) 4.53 × 10−9 m/s
(2) 2.43 × 102 m/s
(3) 1.24 × 108 m/s
(4) 2.21 × 108 m/s

¿Cuál es la velocidad de la luz (f = 5.09×1014 Hz) en alcohol etílico?
(1) 4.53 x 109 m/s
(2) 2.43 x 102 m/s
(3) 1.24 x 108 m/s
(4) 2.21 x 108 m/s

Discussion and Solution:
Remember the formula, , where n1 and n2 = indices of refraction and v1 and v2 = velocities?
We’ll use n1 and v1 to refer to the ethyl alcohol environment and n2 and v2 to refer to a vacuum.
Solving for v1, we get:

v1 = v2

Substituting in what we know, we get:

v1 =  (3.00 x 108 m/s) = 2.21 x 108 m/s

Choice (4).

Problem 2

At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magnitude of the gravitational field strength at this location?
(1) 0.20 kg/N
(2) 5.0 N/kg
(3) 9.8 m/s2
(4) 25 000 N•kg

En una determinada ubicación, una fuerza gravitacional con una magnitud de 350 newtons actúa en un astronauta de 70.-kilogramos. ¿Cuál es la magnitud de la intensidad del campo gravitacional en esta ubicación?
(1) 0.20 kg/N
(2) 5.0 N/kg
(3) 9.8 m/s2
(4) 25 000 N•kg

 Discussion and Solution:
Remember the formula, Fg = mg, where F = force, m = mass, and g = acceleration or gravitational field strength?
Solve the equation for g, we get:

g = Fg / m

Substituting in what we know, we get:

g = Fg / m = 350 N / 70. Kg = 5.0 N / kg

Choice (2).

Problem 3

A spring gains 2.34 joules of elastic potential energy as it is compressed 0.250 meter from its equilibrium position. What is the spring constant of this spring?
(1) 9.36 N/m
(2) 18.7 N/m
(3) 37.4 N/m
(4) 74.9 N/m

Un resorte gana 2.34 joules de energía potencial elástica mientras es comprimido 0.250 metro desde su posición en equilibrio. ¿Cuál es la constante de resorte de este resorte?
(1) 9.36 N/m
(2) 18.7 N/m
(3) 37.4 N/m
(4) 74.9 N/m

Discussion and Solution:
Remember the equation, PEs = ½ kx2, where PEs = potential energy stored in the string, k = spring constant, and x = change in spring length.
Solving the equation for k, we get:

k = 2PEs / x2

Substituting in what we know, we get:

k = 2PEs / x2 = 2(2.34 J) / (0.250 m)2 = 74.9 N / m

Choice (4).

Problem 4

Which statement describes a characteristic common to all electromagnetic waves and mechanical waves?
(1) Both types of waves travel at the same speed.
(2) Both types of waves require a material medium for propagation.
(3) Both types of waves propagate in a vacuum.
(4) Both types of waves transfer energy.

¿Cuál declaración describe una característica en común para todas las ondas electromagnéticas y todas las ondas mecánicas?

  • Ambos tipos de ondas viajan a la misma rapidez.
  • Ambos tipos de ondas requieren un medio material para la propagación.
  • Ambos tipos de ondas se propagan en un vacío.
  • Ambos tipos de ondas transfieren energía.

Discussion and Solution:
Straight memorization, choice (4).

Problem 5

The energy of a sound wave is most closely related to the wave’s
(1) frequency
(2) amplitude
(3) wavelength
(4) speed

La energía de una onda de sonido está relacionada más cercanamente a esta parte de la onda
(1) frecuencia
(2) amplitud
(3) longitud de onda
(4) rapidez

Discussion and Solution:
Straight memorization, choice (2).

 

Chemistry Problems for 9 – 14 – 16

 

Problem 1

Which sample of matter sublimes at room temperature and standard pressure?
(1) Br2()
(2) Cl2(g)
(3) CO2(s)
(4) SO2(aq)

¿Qué muestra de materia realiza sublimación a temperatura ambiente y presión estándar?
(1) Br2(ℓ)
(2) Cl2(g)
(3) CO2(s)
(4) SO2(aq)

Discussion and Solution:
The term sublime refers to when a substance goes from the solid state to the gaseous state without going through the liquid state.
The only solid substance listed is in choice (3).
By The Way, solid carbon dioxide gas is usually referred to as dry ice.

Problem 2

Which reaction occurs at the cathode in an electrochemical cell?
(1) combustion
(2) neutralization
(3) oxidation
(4) reduction

¿Qué reacción ocurre en el cátodo en una celda electroquímica?
(1) combustión
(2) neutralización
(3) oxidación
(4) reducción

Discussion and Solution:
Remember the adage, An Ox, Red Cat?
At the Anode, Oxidation occurs.
Reduction occurs at the Cathode.
Choice (4).

Problem 3

Which substance yields H+(aq) as the only positive ion in an aqueous solution?
(1) CH3CHO
(2) CH3CH2OH
(3) CH3COOH
(4) CH3OCH3

¿Qué sustancia cede H+(aq) como el único ion positivo en una solución acuosa?
(1) CH3CHO
(2) CH3CH2OH
(3) CH3COOH
(4) CH3OCH3

Discussion and Solution:
Only acids release H+ ions in aqueous solution.
The only acid listed is choice (3).
Choice (1) is an aldehyde.
Choice (2) is an alcohol.
Choice (4) is a ketone.

Problem 4

During a nuclear reaction, mass is converted into
(1) charge
(2) energy
(3) isomers
(4) volume

Durante una reacción nuclear, la masa se convierte en
(1) carga
(2) energía
(3) isómeros
(4) volumen

Discussion and Solution:
According to Einstein’s Equation, E = mc2, mass is converted into energy.
Choice (2).

Problem 5

An atom in the ground state has two electrons in its first shell and six electrons in its second shell. What is the total number of protons in the nucleus of this atom?
(1) 5
(2) 2
(3) 7
(4) 8

Un átomo en estado fundamental tiene dos electrones en su primera capa y seis electrones en su segunda capa. ¿Cuál es el número total de protones en el núcleo de este átomo?
(1) 5
(2) 2
(3) 7
(4) 8

Discussion and Solution:
Since the atom is in the ground state, all the electrons are where they are supposed to be.
With a total of eight electrons, in order for the atom to be neutral, it must have eight protons.
Choice (4).

 

 

 

Math Problems for 9 – 12 – 16

 

Problem 1

When factored completely, x3 – 13x2 – 30x is
(1) x(x + 3)(x – 10)
(2) x(x – 3)(x – 10)
(3) x(x + 2)(x – 15)
(4) x(x – 2 )(x + 15)

 Cuando se factoriza por completo, x3 – 13x2 – 30x equivale a
(1) x(x + 3)(x – 10)
(2) x(x – 3)(x – 10)
(3) x(x + 2)(x – 15)
(4) x(x – 2 )(x + 15)

Discussion and Solution:
Since each element in the original equation contains an x, we can factor an x out of each, yielding:

 x(x2 – 13x – 30)

We next need to factor the parenthetical expression, arriving at the final solution.

 x(x2 – 13x – 30) = x(x + 2)(x – 15)

Choice (3).

Problem 2

When (2x – 3)2 is subtracted from 5x2, the result is
(1) x2 – 12x – 9
(2) x2 – 12x + 9
(3) x2 + 12x – 9
(4) x2 + 12x + 9

 Cuando (2x – 3)2 se resta de 5x2, el resultado es
(1) x2 – 12x – 9
(2) x2 – 12x + 9
(3) x2 + 12x – 9
(4) x2 + 12x + 9

Discussion and Solution:
Watch the English. It is asking for 5x2 – (2x – 3)2, not (2x – 3)2 – 5x2.
We first expand (2x – 3)2, getting

5x2 – (4x2 – 12x + 9)

We then negate this because of the negative sign in front of it, getting

5x2 – 4x2 + 12x – 9

We then combine like terms, getting

x2 + 12x – 9

Choice (3).

Problem 3

Which equation is equivalent to y – 34 = x(x – 12)?
(1) y = (x – 17)(x + 2)
(2) y = (x – 17)(x – 2)
(3) y = (x – 6)2 + 2
(4) y = (x – 6)2 – 2

 ¿Qué ecuación es equivalente a y – 34 = x(x – 12)?
(1) y = (x – 17)(x + 2)
(2) y = (x – 17)(x – 2)
(3) y = (x – 6)2 + 2
(4) y = (x – 6)2 – 2

Discussion and Solution:
First we remove the parentheses, getting

y – 34 = x2 – 12x

We next add 36 to both sides, getting

y – 34 + 36 = x2 – 12x + 36

This factors into

y + 2 = (x – 6)2

Finally, we subtract 2 from each side, getting

y = (x – 6)2 – 2

Choice (4).

Problem 4

Which pair of equations could not be used to solve the following equations for x and y?

4x + 2y = 22

– 2x + 2y = – 8

(1) 4x + 2y = 22
2x – 2y = 8

(2) 4x + 2y = 22
– 4x + 4y = – 16

(3) 12x + 6y = 66
6x – 6y = 24

(4) 8x + 4y = 44
– 8x + 8y = – 8

 ¿Qué par de ecuaciones no podría usarse para resolver las siguientes ecuaciones para x e y?

4x + 2y = 22

– 2x + 2y = – 8

(1) 4x + 2y = 22
2x – 2y = 8

(2) 4x + 2y = 22
– 4x + 4y = – 16

(3) 12x + 6y = 66
6x – 6y = 24

(4) 8x + 4y = 44
– 8x + 8y = – 8

Discussion and Solution:
Remember, there are three ways to solve two equations with two unknowns –

  • Graph each equation on the same set of axes. Where they intersect, if they do, is the solution set;
  • Solve one of the equations for one of the unknowns. Then substitute this equality into the other equation and solve. Once the second unknown is determined, substitute this value into either of the original equations to determine the second unknown; and
  • Multiply one (or both) of the two original equations by a small whole number and then add or subtract the two equations, thereby eliminating one of the original unknowns. Substitute this value into either of the two original equations, solving for the second unknown.

From the looks of our choices, we multiply the first equation by 2 and the second equation by 4. When we add these two resulting equations, we would eliminate the value of x.

Choice (4).

Problem 5

Solve the equation for y.

(y – 3)2 = 4y – 12

Resuelva la ecuación para y.

(y – 3)2 = 4y – 12

Discussion and Solution:

First we eliminate the parentheses:

(y – 3)2 = 4y – 12

y2 – 6y + 9 = 4y – 12

We next move all the terms to one side of the equation, getting

y2 – 6y + 9 – 4y + 12 = 0

Combine like terms.

y2 – 10y + 21 = 0

Factor this equation.

(y – 7)(y – 3) = 0

Set each factor equal to 0 and solve for y.

y – 7 = 0

 y = 7

and

y – 3 = 0

 y = 3

 

 

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).

 

Chemistry Problems for 9 – 7 – 16

 

Problem 1

What is the number of electrons in a completely filled second shell of an atom?
(1) 32
(2) 2
(3) 18
(4) 8

¿Cuál es el número de electrones en una segunda capa de un átomo completamente llena?
(1) 32
(2) 2
(3) 18
(4) 8

 Discussion and Solution:
In order to answer this question correctly, you need to have memorized the maximum number of electrons that each shell can hold: 2 – 8 – 18. (Alternatively, many Periodic Tables show these numbers for each element.)
Choice (4).

Problem 2

What is the number of electrons in an atom that has 3 protons and 4 neutrons?
(1) 1
(2) 7
(3) 3
(4) 4

¿Cuál es el número de electrones en un átomo que tiene 3 protones and 4 neutrones?
(1) 1
(2) 7
(3) 3
(4) 4

Discussion and Solution:
Since we are talking about atoms, which are electrically neutral, the number of positive charges (protons) must equal the number of negative charges (electrons).
Choice (3).

Problem 3

As a result of the gold foil experiment, it was concluded that an atom
(1) contains protons, neutrons, and electrons
(2) contains a small, dense nucleus
(3) has positrons and orbitals
(4) is a hard, indivisible sphere

Como resultado de un experimento de láminas de oro, se concluyó que un átomo
(1) contiene protones, neutrones, y electrones
(2) contiene un núcleo pequeño y denso
(3) tiene positrones y orbitales
(4) es una esfera dura e indivisible

Discussion and Solution:
The answer to this question is best determined through straight memorization.
Choice (2).

Problem 4

Which statement describes the distribution of charge in an atom?
(1) A neutral nucleus is surrounded by one or more negatively charged electrons.
(2) A neutral nucleus is surrounded by one or more positively charged electrons.
(3) A positively charged nucleus is surrounded by one or more negatively charged electrons.
(4) A positively charged nucleus is surrounded by one or more positively charged electrons.

¿Qué declaración describe la distribución de cargas en un átomo?
(1) Un núcleo neutral está rodeado por uno o más electrones negativamente cargados.
(2) Un núcleo neutral está rodeado por uno o más electrones positivamente cargados.
(3) Un núcleo positivamente cargado está rodeado por uno o más electrones negativamente cargados.
(4) Un núcleo positivamente cargado está rodeado por uno o más electrones negativamente cargados.

Discussion and Solution:
In a classical atomic description, the nucleus contains positively charged protons, whereas the negatively charged electrons surround the nucleus.
Choice (3).

Problem 5

Which atom, in the ground state, has an outermost electron with the most energy?
(1) Cs
(2) K
(3) Li
(4) Na

¿Qué átomo en estado fundamental tiene un electrón en lo más extremo con la mayor energía?
(1) Cs
(2) K
(3) Li
(4) Na

Discussion and Solution:
Assuming that the electrons are where they normally should be (the ground state), the electrons with the most energy would be those furthest from the nucleus.
The atom that has the most electron shells would show us which one of the choices is correct.
Since Cs is at the bottom of Group 1, it is the correct choice. (1).