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Maslowich
2 years ago
13

ME Project- Decomposer

Physics
1 answer:
Alenkasestr [34]2 years ago
3 0

Answer:

Explanation:

Decomposer.

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How much energy is needed to generate 0.71 x 10-16 kg of mass?
AleksAgata [21]

Answer:

6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass

Explanation:

According to the Equation: E = mc²

where the mass, m = 0.71 * 10⁻¹⁶ kg

the speed of light, c = 3 * 10⁸ m/s

The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:

E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²

E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶

E = 0.71 * 9

E = 6.39 J

6 0
3 years ago
Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
2 years ago
A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in
VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy:<em> This is the energy of a body due to position.</em>

<em>The S.I unit of potential energy is Joules (J).</em>

<em>It can be expressed mathematically as</em>

<em>Ep = mgh........................... Equation 1</em>

<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>

<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

<em />

7 0
2 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
2 years ago
Can you help me with this paper please I will give you 20 points!
sweet [91]
1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

Hope this helps
6 0
2 years ago
Read 2 more answers
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