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Lemur [1.5K]
3 years ago
9

A toy projectile is fired from the ground vertically upward with an initial velocity of 26.5 m/s. The project arrives at its max

imum altitude in 2.7s.
What is the velocity of the projectile when it hits the ground? How?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0
<span>26.5 m/s Ignoring air resistance, the projectile will have an initial kinetic energy expressed by E = 0.5 M V^2 and as the projectile travels upward, it's kinetic energy will decrease while it's potential gravitational kinetic energy increases such that the sum of both is constant. Then once it starts to descent, the potential energy is converted back into kinetic energy. And when it finally reaches the ground, it's kinetic energy will exactly match the original kinetic energy it had initially. And the only way it can match since the mass never changes is for it to have the same velocity as it had when it was first fired.</span>
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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

       

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3 years ago
A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S
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zlopas [31]

Answer:

330.5  m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

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g= 10 m/s²

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