All categories

2 years ago

A ray diagram is shown. Which statement best describes the diagram?

1 answer:

6 0

The statement that best describes the diagram is:

"The diagram is destined to remain forever
an enigma wrapped in a mystery, since its
willing audience was denied the merest peek."

You might be interested in

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

8 0

1 year ago

Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo

lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0

2 years ago