Use this free online force converter to change joules per centimeter into femtonewtons instantly. Type in the joules per centimeter value, and the equivalent femtonewtons 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
Femtonewtons
How to use this Joules per Centimeter to Femtonewtons Converter 🤔
Follow these steps to convert given Joules per Centimeter value from Joules per Centimeter units to Femtonewtons units.
Enter the input Joules per Centimeter value in the text field.
The given Joules per Centimeter is converted to Femtonewtons in realtime ⌚ using the formula, and displayed under the Femtonewtons label.
You may copy the resulting Femtonewtons value using the Copy button.
Formula
To convert given force from Joules per Centimeter to Femtonewtons, use the following formula.
Femtonewtons = Joules per Centimeter * 1e+13
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.
Femtonewtons
A femtonewton (fN) is one-quadrillionth of a newton. It is used in experimental physics and nanotechnology to measure forces at the atomic scale, such as the forces involved in the interaction between nanoparticles.
{
"conversion": "joule-per-centimeter-femtonewton",
"x_slug": "joule-per-centimeter",
"y_slug": "femtonewton",
"x": "J/cm",
"y": "fN",
"x_desc": "Joules per Centimeter",
"y_desc": "Femtonewtons",
"category": "Force",
"symbol": "m",
"formula": "x * 1e+13",
"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 Femtonewtons.</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 femtonewtons is:</p>\n <p class=\"formula step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>Force<sub>(Joules per Centimeter)</sub></span> × 1e+13</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>(Femtonewtons)</sub></span> = <span>500</span> × 1e+13</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = 5000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>500 J/cm</strong> is equal to <strong>5000000000000000 fN</strong>.</p>\n <p>The force is <strong>5000000000000000 fN</strong>, in femtonewtons.</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 Femtonewtons.</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 femtonewtons is:</p>\n <p class=\"formula step\"><span>Force<sub>(Femtonewtons)</sub></span> = <span>Force<sub>(Joules per Centimeter)</sub></span> × 1e+13</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>(Femtonewtons)</sub></span> = <span>1000</span> × 1e+13</p>\n <p class=\"step\"><span>Force<sub>(Femtonewtons)</sub></span> = 10000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>1000 J/cm</strong> is equal to <strong>10000000000000000 fN</strong>.</p>\n <p>The force of machine is <strong>10000000000000000 fN</strong>, in femtonewtons.</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": "A femtonewton (fN) is one-quadrillionth of a newton. It is used in experimental physics and nanotechnology to measure forces at the atomic scale, such as the forces involved in the interaction between nanoparticles.",
"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."
}
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