Use this free online force converter to change joules per centimeter into joules per meter instantly. Type in the joules per centimeter value, and the equivalent joules per meter is calculated for you in real time.
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Enter your inputs, and the result is calculated in real-time.
Joules per Centimeter
Joules per Meter
How to use this Joules per Centimeter to Joules per Meter Converter 🤔
Follow these steps to convert given Joules per Centimeter value from Joules per Centimeter units to Joules per Meter units.
Enter the input Joules per Centimeter value in the text field.
The given Joules per Centimeter is converted to Joules per Meter in realtime ⌚ using the formula, and displayed under the Joules per Meter label.
You may copy the resulting Joules per Meter value using the Copy button.
Formula
To convert given force from Joules per Centimeter to Joules per Meter, use the following formula.
Joules per Meter = Joules per Centimeter * 0.01
Calculation
Calculation will be done after you enter a valid input.
Joules per Centimeter
Joule per centimeter (J/cm) is similar to joule per meter but is used when a finer unit of measurement is needed. It is also used to describe surface energy and tension in materials science.
Joules per Meter
Joule per meter (J/m) is a unit that represents energy per unit length. It is often used in the context of surface tension, where it describes the energy required to increase the surface area of a liquid.
{
"conversion": "joule-per-centimeter-joule-per-meter",
"x_slug": "joule-per-centimeter",
"y_slug": "joule-per-meter",
"x": "J/cm",
"y": "J/m",
"x_desc": "Joules per Centimeter",
"y_desc": "Joules per Meter",
"category": "Force",
"symbol": "m",
"formula": "x * 0.01",
"examples": "<div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">1</span>\n <h3 class=\"question\">Consider a force applied over 200 centimeters with an energy of 500 joules per centimeter.<br>Convert this force from joule-per-centimeter to Joules per Meter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force in joules per centimeter is:</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 500</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from joules per centimeter to joules per meter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>Force<sub>(Joules per Centimeter)</sub></span> × 0.01</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight <strong>Force<sub>(Joules per Centimeter)</sub> = 500</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>500</span> × 0.01</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 5</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>500 J/cm</strong> is equal to <strong>5 J/m</strong>.</p>\n <p>The force is <strong>5 J/m</strong>, in joules per meter.</p>\n </div>\n <div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">2</span>\n <h3 class=\"question\">Consider a machine that exerts 1,000 joules per centimeter while compressing material.<br>Convert this force from joule-per-centimeter to Joules per Meter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of machine in joules per centimeter is:</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 1000</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from joules per centimeter to joules per meter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>Force<sub>(Joules per Centimeter)</sub></span> × 0.01</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of machine, <strong>Force<sub>(Joules per Centimeter)</sub> = 1000</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = <span>1000</span> × 0.01</p>\n <p class=\"step\"><span>Force<sub>(Joules per Meter)</sub></span> = 10</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>1000 J/cm</strong> is equal to <strong>10 J/m</strong>.</p>\n <p>The force of machine is <strong>10 J/m</strong>, in joules per meter.</p>\n </div>\n ",
"units": [
[
"newton",
"Newtons",
"N"
],
[
"kilonewton",
"Kilonewtons",
"kN"
],
[
"gram-force",
"Gram-Force",
"gf"
],
[
"kilogram-force",
"Kilogram-Force",
"kgf"
],
[
"ton-force",
"Metric Ton-Force",
"tf"
],
[
"exanewton",
"Exanewtons",
"EN"
],
[
"petanewton",
"Petanewtons",
"PT"
],
[
"teranewton",
"Teranewtons",
"TN"
],
[
"giganewton",
"Giganewtons",
"GN"
],
[
"meganewton",
"Meganewtons",
"MN"
],
[
"hectonewton",
"Hectonewtons",
"hN"
],
[
"dekanewton",
"Dekanewtons",
"daN"
],
[
"decinewton",
"Decinewtons",
"dN"
],
[
"centinewton",
"Centinewtons",
"cN"
],
[
"millinewton",
"Millinewtons",
"mN"
],
[
"micronewton",
"Micronewtons",
"µN"
],
[
"nanonewton",
"Nanonewtons",
"nN"
],
[
"piconewton",
"Piconewtons",
"pN"
],
[
"femtonewton",
"Femtonewtons",
"fN"
],
[
"attonewton",
"Attonewtons",
"aN"
],
[
"dyne",
"Dynes",
"dyn"
],
[
"joule-per-meter",
"Joules per Meter",
"J/m"
],
[
"joule-per-centimeter",
"Joules per Centimeter",
"J/cm"
],
[
"ton-force-short",
"Short Ton-Force",
"short tonf"
],
[
"to-force-long",
"Long Ton-Force (UK)",
"tonf (UK)"
],
[
"kip-force",
"Kip-Force",
"kipf"
],
[
"kilopound-force",
"Kilopound-Force",
"kipf"
],
[
"pound-force",
"Pound-Force",
"lbf"
],
[
"ounce-force",
"Ounce-Force",
"ozf"
],
[
"poundal",
"Poundals",
"pdl"
],
[
"pound-foot-per-square-second",
"Pound Foot per Square Second",
"lbf·ft/s²"
],
[
"pond",
"Ponds",
"p"
],
[
"kilopond",
"Kiloponds",
"kp"
]
],
"y_long_desc": "Joule per meter (J/m) is a unit that represents energy per unit length. It is often used in the context of surface tension, where it describes the energy required to increase the surface area of a liquid.",
"x_long_desc": "Joule per centimeter (J/cm) is similar to joule per meter but is used when a finer unit of measurement is needed. It is also used to describe surface energy and tension in materials science."
}