Use this free online force converter to change giganewtons into joules per centimeter instantly. Type in the giganewtons value, and the equivalent joules per centimeter is calculated for you in real time.
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Giganewtons
Joules per Centimeter
How to use this Giganewtons to Joules per Centimeter Converter 🤔
Follow these steps to convert given Giganewtons value from Giganewtons units to Joules per Centimeter units.
Enter the input Giganewtons value in the text field.
The given Giganewtons is converted to Joules per Centimeter in realtime ⌚ using the formula, and displayed under the Joules per Centimeter label.
You may copy the resulting Joules per Centimeter value using the Copy button.
Formula
To convert given force from Giganewtons to Joules per Centimeter, use the following formula.
Joules per Centimeter = Giganewtons * 1e+11
Calculation
Calculation will be done after you enter a valid input.
Giganewtons
A giganewton (GN) is 109 newtons. It’s used to describe extremely large forces, such as the thrust produced by spacecraft or the force involved in geological phenomena like earthquakes. Giganewtons help put into perspective the vast power involved in significant natural or artificial forces.
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.
{
"conversion": "giganewton-joule-per-centimeter",
"x_slug": "giganewton",
"y_slug": "joule-per-centimeter",
"x": "GN",
"y": "J/cm",
"x_desc": "Giganewtons",
"y_desc": "Joules per Centimeter",
"category": "Force",
"symbol": "m",
"formula": "x * 1e+11",
"examples": "<div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">1</span>\n <h3 class=\"question\">Consider a large rocket engine producing a thrust of 100 giganewtons.<br>Convert this thrust from giganewtons to Joules per Centimeter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of rocket engine in giganewtons is:</p>\n <p class=\"step\"><span>Force<sub>(Giganewtons)</sub></span> = 100</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from giganewtons to joules per centimeter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>Force<sub>(Giganewtons)</sub></span> × 1e+11</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of rocket engine, <strong>Force<sub>(Giganewtons)</sub> = 100</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>100</span> × 1e+11</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 10000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>100 GN</strong> is equal to <strong>10000000000000 J/cm</strong>.</p>\n <p>The force of rocket engine is <strong>10000000000000 J/cm</strong>, in joules per centimeter.</p>\n </div>\n <div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">2</span>\n <h3 class=\"question\">Consider a superstructure experiencing 50 giganewtons of force.<br>Convert this force from giganewtons to Joules per Centimeter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of superstructure in giganewtons is:</p>\n <p class=\"step\"><span>Force<sub>(Giganewtons)</sub></span> = 50</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from giganewtons to joules per centimeter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>Force<sub>(Giganewtons)</sub></span> × 1e+11</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of superstructure, <strong>Force<sub>(Giganewtons)</sub> = 50</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>50</span> × 1e+11</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 5000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>50 GN</strong> is equal to <strong>5000000000000 J/cm</strong>.</p>\n <p>The force of superstructure is <strong>5000000000000 J/cm</strong>, in joules per centimeter.</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"
]
],
"x_long_desc": "A giganewton (GN) is 10<sup>9</sup> newtons. It’s used to describe extremely large forces, such as the thrust produced by spacecraft or the force involved in geological phenomena like earthquakes. Giganewtons help put into perspective the vast power involved in significant natural or artificial forces.",
"y_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."
}
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