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Vilka [71]
3 years ago
10

How much heat is required to heat 0.1 g of ∆hvap =2260 j/g ∆h =340j/g fus iceat−30ctosteamat100c? use the approximate values?

Physics
1 answer:
Anestetic [448]3 years ago
7 0
How much heat<span> is </span>required<span> to </span>heat 0.1 g<span> of </span>∆hvap<span> =</span>2260 j/g ∆h<span> =</span>340j/g fus iceat−30 ctosteamat 100c?use<span> the </span>approximate values<span>?</span>
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Identify the equation used to calculate the perpendicular force (F⊥) acting on a block on an inclined plane.
pychu [463]
Using geometrical arguments, we can see that the angle of the inclined plane \theta is equal to the angle between Fg and the perpendicular force.

But the perpendicular force is the projection of Fg along the perpendicular axis, and Fg=mg, so the correct answer is
<span>C) F=mg cosΘ </span>
6 0
3 years ago
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How does Scientific theories differ from scientific laws
finlep [7]
Hey there!

\ggg \ scientific \ laws : Scientific law's would be a law that was proven and test and examined by scientist.These laws would basically be fact's, proven that what ever they say would technically be true.

\ggg \ Scientific \ theories: These are "theories" that are made by scientist usually hypothesis to see what law would actually be true. These "theories" are of course not true, they are not quite laws, they are experiment's that could be laws, but they're theories, thing's that are technically false.

Hope this helps you!
7 0
4 years ago
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8 0
4 years ago
A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55
svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate  =  2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v    ....................1

h = v²/2g   ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0
4 years ago
If you have 4 kg of a sample with density of 1 897 g/ml what is the volume
Kaylis [27]
We know, Volume = Mass / Density
Here, mass = 4 Kg = 4000 g
d = 1,897 g/ml

Substitute their values, 
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v = 2.108 ml

In short, Your Answer would be 2.108 mL

Hope this helps!
3 0
3 years ago
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