Avogadro's law states that at a given temperature and pressure, what quantity is constant?

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

The ammonia molecule (NH3) has a dipole moment of 5.0×10?30C?m. Ammonia molecules in the gas phase are placed in a uniform elect

Neko [114]

Question (continuation)

(a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E S from parallel to perpendicular?

(b) At what absolute temperature T is the average translational kinetic energy 3/2kT of a molecule equal to the change in potential energy calculated in part (a)?

Answer:

a. 9.0 * 10^-24 Joules

b. 0.44K

Explanation:

Given

Let p = dipole moment = 5.0 * 10^-30 Cm

Let E = Magnitude = 1.8 * 10^6 N/m

a.

The charge in electric potential = Final Charge - Initial Charge

Initial Charge = Potential Energy

Initial Energy = -pE cosФ where Ф = 0

So, initial Energy = - 5.0 * 10^-30 * 1.8 * 10^6

Initial Energy = -9 * 10^-24 Joules

Final Energy = 0

Charge = 0 - (-9.0 * 10^-24)

Charge = 9.0 * 10^-24 Joules

b.

Absolute Temperature

Change in Kinetic Energy = Change in Potential Energy = 9.0 * 10^-24

Change in Kinetic Energy = 3/2kT where k is Steven-Boltzmann constant = 1.38 * 10^-23

So,

9.0 * 10^-24 = 3/2 * 1.38 * 10^-23 * T

T = (9.0 * 10^-24 * 2)/(3 * 1.38 * 10^-23)

T = (18 * 10^-24)/(4.14 * 10^-23)

T = 0.44K

6 0

3 years ago

What is the most commonly used reference point

snow_tiger [21]

Answer:

El punto de referencia más usado es el NORTE. Los puntos de referencia sirven para orientarnos, de esta manera podemos saber nuestra ubicación y luego desplazarnos. En este caso tenemos que el punto cardinal NORTE es el más usado, se tiene como referencia universal para lograr ubicarnos.

Explanation:

gracias

7 0

3 years ago

Read 2 more answers

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

How does water get up to the atmosphere, and how does it get back down to earth surface

IrinaK [193]

Answer:

Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).

6 0

3 years ago