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Black_prince [1.1K]
3 years ago
13

How much heat in joules is required to heat a 50 g sample of aluminum from 71 ∘f to 142 ∘f?

Physics
1 answer:
Vsevolod [243]3 years ago
3 0
A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
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stepan [7]
All Mountains are built through a general process called "deformation" of the crust of the Earth. Deformation is a fancy word which could also mean "folding". An example of this kind of folding comes from the process described below. 

<span>When two sections of the Earth's lithosphere collide, rather than being subducted, where one slab of lithosphere is forced down to deeper regions of the Earth, the slabs pile into each other, causing one or both slabs can fold up like an accordion. This process elevates the crust, folds and deforms it heavily, and produces a mountain range. Mountain building and mantle subduction usually go together. </span>
3 0
3 years ago
The normal boiling point of a certain liquid is , but when of urea () are dissolved in of the solution boils at instead. Use thi
Kruka [31]

Answer:

100 Degrees is boiling point.

Explanation:

5 0
3 years ago
Why is Earth’s North Pole a geographic north pole but a south seeking pole magnetically?
natima [27]

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3 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
Calculate the acceleration of a 90-kg skydiver with a net force of -883 N
Helen [10]
F=ma, in this case your force=-883 N and mass = 90 kg. F/m=a therefore acceleration=-883/90=-9.81 m/s/s
6 0
3 years ago
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