The illuminance on the surface can be calculated bu using the formula:
E=I/s2. E stands for the illuminance on the surface (in lux), I stands for theintensity of the light from the source (in candelas), and s stands for the distance from the light source to the surface (in meters).
So, the answer is: 8 lux
8 lux.
I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.
The amount of separation between two horizontal surface/Plane
One factor is the height of the ramp. The higher the height of the ramp the further the car travels. Another factor is the surface of the ramp. With a rough surface on the ramp e.g sand paper the car travels a short distance. With a lubricated surface on the ramp e.g Vaseline the car will travel a very long distance.
Surface tension.
8 lux.
a. 8 lux b. 24 lux c. 216 lux d. 648 lux
On the surface
Illuminance is another term for luminous-flux density which defines the intensity of the luminous flux arriving at a surface, measured in lumens per square metre, which is given the special name 'lux'.
The most commonly used metric for measuring daylight in a building is called the "daylight factor"-expressed as a percent of available exterior daylight.Daylight factor is defined as a ratio of interior illuminance at a horizontal point in a building to the sun-shaded exterior illuminance on a horizontal surface under a fully overcast sky (a ratio of outside illuminance over inside illuminance).The higher the daylight factor, the more natural light that is available in the room.
A footcandle is a unit of measurement for illuminance on a surface, equal to one lumen per square foot. A footlambert, on the other hand, is a unit of luminance, measuring the brightness of a surface emitting or reflecting light in footlamberts.
Illuminance is another term for luminous-flux density which defines the intensity of the luminous flux arriving at a surface, measured in lumens per square metre, which is given the special name 'lux'.
On the surface
Lux is unit for measuring the illumination (illuminance) of a surface. It is the amount of light received per unit of surface area. You can' t convert watt, which is power, to Lux.
Speed and distance
I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.
There is no fixed distance of the ozone from surface. It is said to be an approximate of 15 to 55 km.