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).

 

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