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2 years ago

Donde esta la orilla del universo?

Physics

1 answer:

irinina [24]2 years ago

5 0

Answer:

<em>El universo no tiene un borde finito</em>, continúa y se expande sin fin. La distancia adecuada desde la Tierra hasta el borde del universo observable es de aproximadamente 46.5 mil millones de años luz o 4.40 × 10 ^ 26 metros en cualquier dirección.

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What is the equation for determining the value of a specific gas constant using the universal gas constants

maria [59]

Answer:

Physically, the gas constant is the constant of proportionality that relates the energy scale in physics to the temperature scale, when a mole of particles at the stated temperature is being considered. Thus, the value of the gas constant ultimately derives from historical decisions and accidents in the setting of the energy and temperature scales, plus similar historical setting of the value of the molar scale used for the counting of particles.

Explanation:

Pa follow

3 0

2 years ago

Two sound waves, A and B, are traveling at the same speed. Wave A has a wavelength of 50 cm and a frequency of 7000 Hz. Wave B h

vekshin1

Answer:

D. 100 cm

Explanation:

The speed of a wave is the wavelength times the frequency.

v = λf

Wave A and B have the same speed, so:

λf = λf

(50 cm) (7000 Hz) = λ (3500 Hz)

λ = 100 cm

6 0

2 years ago

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

3 years ago

A curve in a stretch of highway has radius R. The road is not banked in any way. The coefficient of static friction between the

adelina 88 [10]

Answer:

maximum possible velocity = $\sqrt{ugR}$

Explanation:

centripetal acceleration when the car is going in the circle must be less than the maximum friction for the car to not slip.

centripetal acceleration $\frac{mv^{2}}{r}$

where v is the velocity of car and r is the radius of circle

maximum friction = umg

where u is the coefficient of static friction.

therefore $umg\geq \frac{mv^{2}}{R}$

therefore maximum possible velocity = $\sqrt{ugR}$

6 0

3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago