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lisabon 2012 [21]
2 years ago
6

As your rockets went upwards how would you describe how the energies changed?

Physics
1 answer:
chubhunter [2.5K]2 years ago
4 0

As a rocket increases height and slows down, it gains more and more potential energy and loses more and more kinetic energy. Potential energy is store energy (usually determined by height), and kinetic energy increases as speed increases.

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This should be pretty easy i think.
timama [110]

Answer:

its a sure for not but i think its a 192583

3 0
2 years ago
According to Newton's Second Law of Motion, an object will accelerate if you apply what kind of force? Question 1 options: Frict
Bumek [7]

An unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

<h3>What does Newton's Second Law of Motion state?</h3>

It states that the force applied to the object is equal to the product of mass and acceleration.

F = ma

  • An object will accelerate when the net force applied on the object is more than zero or unbalanced.
  • The acceleration is the change in the direction or speed of the object. To achieve acceleration the force must be greater in a direction.

  • When force is greater in one the object move in that direction which is known as acceleration.

Therefore, an unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

Learn more about Newton's Second Law of Motion.:

brainly.com/question/25810165

5 0
2 years ago
Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou
olga2289 [7]

The conclusion that is best supported by the data is;

D) A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

3 0
3 years ago
Read 2 more answers
How does the currently accepted model of the nucleus provide evidence of the existence of the strong nuclear force?
olganol [36]

Answer:

Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.

4 0
3 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
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