An ion has 8 protons and 10 electrons. What is the total charge? *

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

8 0

3 years ago

Anything that has volume or mass

irinina [24]

Any object, except antimatter, :)

8 0

3 years ago

A meter stick hurtles through space at a speed of 0. 95c relative to you, with its length perpendicular to the direction of moti

marta [7]

Answer:

Explanation:

Caty , Use the relativity formula for length. ( they teach this in H.S. ? ) it's from my Modern Physics in college, A 300 level class

L = $L_{0}$ $\sqrt{1-\frac{v^{2} }{c^{2} } }$

L = 3 $\sqrt{1-\frac{.95^{2} }{1^{2} } }$

L = 0.9367496998 meters

L = 0.94 meters approx

6 0

2 years ago

A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati

Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π $\sqrt{\frac{l}{g} }$

angular frequency ω = 2π /T

= $\sqrt{\frac{g}{l} }$

ω₁ /ω₂ = $\sqrt{\frac{g_1}{g_2} }$

Putting the values

ω₁ = ω , ω₂ = ω / 4

ω₁ /ω₂ = 4

4 = $\sqrt{\frac{g}{g_2} }$

g₂ = g / 16

option d is correct.

6 0

3 years ago