Use this free online force converter to change petanewtons into joules per centimeter instantly. Type in the petanewtons value, and the equivalent joules per centimeter is calculated for you in real time.
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Petanewtons
Joules per Centimeter
How to use this Petanewtons to Joules per Centimeter Converter 🤔
Follow these steps to convert given Petanewtons value from Petanewtons units to Joules per Centimeter units.
Enter the input Petanewtons value in the text field.
The given Petanewtons 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 Petanewtons to Joules per Centimeter, use the following formula.
Joules per Centimeter = Petanewtons * 1e+17
Calculation
Calculation will be done after you enter a valid input.
Petanewtons
A petanewton (PT) is equal to 1015 newtons. It’s an extremely large unit of force, mostly theoretical, used in astrophysics or high-energy physics to describe forces acting at a massive scale, such as between planets or in star formation.
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": "petanewton-joule-per-centimeter",
"x_slug": "petanewton",
"y_slug": "joule-per-centimeter",
"x": "PT",
"y": "J/cm",
"x_desc": "Petanewtons",
"y_desc": "Joules per Centimeter",
"category": "Force",
"symbol": "m",
"formula": "x * 1e+17",
"examples": "<div class=\"example\">\n <div class=\"example_head\"><span class=\"example_n\">1</span>\n <h3 class=\"question\">Consider a planet exerting a force of 10 petanewtons on its moon.<br>Convert this force from petanewtons to Joules per Centimeter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of planet in petanewtons is:</p>\n <p class=\"step\"><span>Force<sub>(Petanewtons)</sub></span> = 10</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from petanewtons to joules per centimeter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>Force<sub>(Petanewtons)</sub></span> × 1e+17</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of planet, <strong>Force<sub>(Petanewtons)</sub> = 10</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>10</span> × 1e+17</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 1000000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>10 PT</strong> is equal to <strong>1000000000000000000 J/cm</strong>.</p>\n <p>The force of planet is <strong>1000000000000000000 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 the energy needed for interstellar travel requiring a force of 0.8 petanewtons.<br>Convert this force from petanewtons to Joules per Centimeter.</h3></div>\n <h4 class=\"answer\">Answer:</h4>\n <p><strong>Given:</strong></p>\n <p>The force of interstellar vehicle required in petanewtons is:</p>\n <p class=\"step\"><span>Force<sub>(Petanewtons)</sub></span> = 0.8</p>\n <p><strong>Formula:</strong></p>\n <p>The formula to convert force from petanewtons to joules per centimeter is:</p>\n <p class=\"formula step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>Force<sub>(Petanewtons)</sub></span> × 1e+17</p>\n <p><strong>Substitution:</strong></p>\n <p>Substitute given weight of interstellar vehicle required, <strong>Force<sub>(Petanewtons)</sub> = 0.8</strong> in the above formula.</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = <span>0.8</span> × 1e+17</p>\n <p class=\"step\"><span>Force<sub>(Joules per Centimeter)</sub></span> = 80000000000000000</p>\n <p><strong>Final Answer:</strong></p>\n <p>Therefore, <strong>0.8 PT</strong> is equal to <strong>80000000000000000 J/cm</strong>.</p>\n <p>The force of interstellar vehicle required is <strong>80000000000000000 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 petanewton (PT) is equal to 10<sup>15</sup> newtons. It’s an extremely large unit of force, mostly theoretical, used in astrophysics or high-energy physics to describe forces acting at a massive scale, such as between planets or in star formation.",
"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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