Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
Verrrrry interesting !
If the moon were replaced by something with a vastly greater mass
but at the same distance, then ...
-- The period of its revolution around the Earth would be much shorter.
That is, it would orbit the Earth in much less than 27.3 days. We might
see it go through a complete set of phases in 2 weeks, or even 1 week.
-- The ocean tides would be much greater. Low tides would be
much lower, and high tides would be much higher.
-- Sadly, the land tides, and the forces on the Earth's internal structure,
would also be much greater. That means great increases in earthquake
and volcanic activity.
-- The Earth and moon both revolve around their common center of
mass. Under the current arrangement ... with the Earth having 80 times
the mass of the Moon ... that point is inside the Earth, and it looks a lot
like the Moon is orbiting a stationary Earth.
When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it.
Then, it will look a lot like the monster is the stationary one, and the
Earth is orbiting it.
I actually don't believe that we would SEE that change, or feel it.
Answer:
3400 m
Explanation:
Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;
speed = distance/time, hence, distance = speed x time.
<em>For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as</em>;
distance = 340 m/s x 10 s = 3400 m
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
brainly.com/question/14548124