The energy principle states that:

Energy can be destroyed.
Energy can be created.
Energy cannot be created or destroyed.
Energy cannot be created, but it can be destroyed.


brainleist to correct answer

Answers

Answer 1
Energy cannot be created or destroyed

I hope that helps
Answer 2

The energy principle states that:

[tex]\sf\purple{Energy \:cannot \:be \:created \:or\: destroyed.✅}[/tex]

Law of conservation of energy or the first law of thermodynamics states that energy can neither be created nor destroyed; it can only be transferred or changed from one form to another.

[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{.}}}}}[/tex]


Related Questions

Which is a valid ionic compound?

sodide chlorine
sodium chlorine
sodium chloride
sodide chloride

Answers

Answer:

sodium chloride

Explanation:

Sodium chloride is an ionic compound. The ions of the sodium chloride compound is sodium ion and chloride ion.  The sodium ion is cation while the chloride ion is an anion. Sodium chloride is a very stable compound because of the mutual attraction of oppositely charged ions.

Transformar las siguientes unidades al Sistema Internacional: 30 km/h ; 37 Dm ; 750 g ; 4x10-6 km2 ; 7500 cm ; 600000 cm2 ; 520700000 mm3 ; 3,4 años.

Answers

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Which statement best describes the Sun Earth-Moon system?

Answers

Answer:

A

Explanation:

A. The Moon orbits the Earth, and the Earth-Moon system orbits the Sun

B. The Earth and Moon both orbit the Sun separately

C. The Sun orbits the Earth-Moon system

D. The Moon revolves around Earth but not the Sun

Which object will take the most force
to accelerate? *
4 kg
6 kg
8 kg
02 kg

Answers

Answer:

I think it might be 8kg grams because it is bigger

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.

Answers

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

 

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La resistividad de cierto alambre es 1.72x10 -8 Ωm a 20 oC .Una batería de 6v se conecta a una bobina de 20 m de este alambre, con un diámetro de 0.8 mm. ¿Cuál es la corriente en el alambre?

Answers

Answer:

   i = 1.09 10⁴ A

Explanation:

For this exercise we will look for the resistance of the wire

          R = ρ L / a

the area of ​​the wire is

          a =ππ r² = π πd² / 4

we substitute

         R = ρ L 4 / π d²

          R = 1.72 10⁻⁸   20  4/π 8 10⁻⁴

         

         R = 5.47 10⁻⁴ Ω

to calculate the current we use ohm's law

           V = R i

            i = V / R

            i = 6 / 5.47 10⁻⁴

            i = 1.09 10⁴ A

You have just learned that the planet core of Mars is no longer radiating heat like the core of Earth. With this information what would you communicate to building designers as the lead geologist in regards to suitable building locations on the surface of Mars? Remember your informing builders about locations that would best avoid natural disasters like marsquakes and volcanoes? What evidence or theories do you have to support your statements?

Answers

Answer:

Explanation:

That is a fun question!

Without a hot radiating core like Earth, Mars will have very different geological and seismic events. The Mars core will be relatively cold and there will not be any molten magma. So Mars will not have earthquakes or volcano activities. Both only occur when there is magma flowing or tectonic plate motion and they will not occur with a cold core.

Pop up spring Lab
Describe the energy conversion that took place. What energy was present just before the jump? What energy is present at its maximum height? What happens in between?
Draw bar charts of energy before the jump, while the spring toy is compressed and after the jump at the maximum height of its jump. Be sure to label each type of energy.

Answers

Answer:

I honestly don't know but that thing looks cool lol

Physicist ____ was found to have completely made up some of the data on which his astounding discoveries were made.

Answers

Answer:

Jan Hendrik Schön I believe

Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges become 2q and the distance becomes 3r, what are the new mutual forces

Answers

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

Give your answer to 2 dp
When taking off a plane accelerates at 2.7m/s2 down the runway. It accelerates from a stationary position for 25 seconds before leaving the ground. What
is the planes speed when it leaves the ground?​

Answers

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

Explain how frequency and amplitude affect sound vibrations?

Answers

Answer:

The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.

Explanation:

A bird in flight pushes itself upward with
a 7.28 N force. If the bird is climbing at a
constant rate of 1 m/s (no acceleration),
what is the weight of the bird?
[?] N

Answers

Answer:

The weight of the bird is equal to 7.28 N.

Explanation:

The upward force acting on the bird = 7.28 N

The bird is climbing at a constant rate of 1 m/s.

We need to find the weight of the bird.

We know that the weight of an object is the force of gravity acting on it. It can be calculated as follows :

W = mg

In this case, 7.28 N of force is acting on the object. Hence, the weight of the bird is equal to 7.28 N.

What are the si units

Answers

Answer:

The uniy which is accepted all over the world is called SI unit.

Explanation:

The system of measurement that is agreed by the international convention if scientists that is held in paris of France to adopt an international unit is called SI unit unit.

A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.

Answers

Answer:

3.504 MeV

Explanation:

Given that;[tex]\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q[/tex]

Also;

[tex]Q= KE_{\alpha } (M_{Y} + M_{\alpha } /M_{Y[/tex])

Mass number of X = 229

Mass number of Y = 225

Mass number of alpha particles = 4

Kinetic energy of alpha particles = 3.443 MeV

Q = 3.443 MeV (225 + 4/225)

Q= 3.504 MeV

An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What is the length of the rod at the freezing point of water (0 0C)? (b) What is the temperature if the length of the rod is 10.009 cm? /4mks

Answers

Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

A charge Q acts on a point charge to create an electric field. Its strength, measured a distance of 40 cm away is 100 N/C. What is the magnitude of the electric field strength at a distance of 20 cm

Answers

Answer:

[tex]E_2=80N/C[/tex]

Explanation:

From the question we are told that:

Initial Distance [tex]d_1=40cm=>0.4m[/tex]

Initial Electric field strength [tex]E_1=100N/C[/tex]

Final Distance [tex]d_2=20cm=>0.20m[/tex]

Generally the equation for Electric field is mathematically given by.

 [tex]E=\frac{kq}{d^2}[/tex]

 [tex]q=\frac{100*(0.4)^2}{K}[/tex]

Substituting q for [tex]d=20cm[/tex]

 [tex]E_2=\frac{k}{0.2}*\frac{100*(0.4)^2}{K}[/tex]

 [tex]E_2=80N/C[/tex]

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