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spin [16.1K]
3 years ago
8

What is the difference between potential energy and kinetic energy?

Physics
1 answer:
gavmur [86]3 years ago
6 0
The answer i believe is A hope it helps
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A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f
Verdich [7]

Answer:

\Delta T = 0.81 ^oC

Explanation:

As we know by energy conservation

All its kinetic energy will convert into thermal energy to raise its temperature

\frac{1}{2}mv^2 = ms\Delta T

now divide both sides by mass of the object

\frac{1}{2}v^2 = s\Delta T

so change in temperature is given as

\Delta T = \frac{v^2}{2s}

\Delta T = \frac{25^2}{2\times 387}

\Delta T = 0.81^oC

3 0
3 years ago
How much clothes should be applied on 100 cm² area using pressure of 25 Pa
IrinaVladis [17]

Answer:

hi !

Pressure = Force/Area

Force = Pressure*Area

Pressure = 25 Pa

Area = 100 cm² = 0.01 m²

F = 25*0.01

   = 0.25 N

Explanation:

6 0
3 years ago
Which of the mathematical sign below would best complete the following statement?
vredina [299]

Answer:

B

Explanation:

option b stands for less down and cm is less than km

6 0
3 years ago
1. Determine the energy released per kilogram of fuel used.
yarga [219]

200 MeV of energy  
E1/E2=7.61=8

U is equal to 1 kilogram or 1000 g.
There are 6.02310 23 atoms in one mole, or 235 g, of uranium. Therefore, 6.02310 23 atoms are present in 1000 g of 92/235 U.
It is understood that one atom releases 200 MeV of energy during its fission.

As a result, the energy released from the fission of one kilogram of 92/235 is given by E 2 = 6.02310 23 1000200/235 =5.10610 26 MeV E1/E2=7.61=8

In light of this, the energy released during the fusion of one kilogram of hydrogen is roughly eight times greater than the energy generated during the fission of one kilogram of uranium.

To learn more about Fission please visit -
brainly.com/question/27923750
#SPJ1

6 0
2 years ago
How much energy is needed to evaporate 5 grams of water?
ella [17]
Explanation:
For the reaction.......
H
2
O
(
l
)
+
Δ
→
H
2
O
(
g
)
WHERE BOTH PRODUCT AND REACTANT ARE AT
100

∘
C
,
..........
Δ
H
∘
vaporization
=
40.66
⋅
k
J
⋅
m
o
l
−
1
And thus we need to assess the molar quantity of the water vaporized......
Δ
H
rxn
=
5.00
⋅
g
18.01
⋅
g
⋅
m
o
l
−
1
×
40.66
⋅
k
J
⋅
m
o
l
−
1
=11.3 kJ

6 0
3 years ago
Read 2 more answers
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