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Anni [7]
3 years ago
6

7. A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its

original speed. What is the mass of the second ball
Physics
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer:

mass of the second ball is 0.379m

Explanation:

Given;

mass of first ball = m

let initial velocity of first ball = u₁

let final velocity of first ball  = v₁ = 0.45u₁

let the mass of the second ball = m₂

initial velocity of the second ball, u₂ = 0

let the final velocity of the second ball = v₂

Apply the principle of conservation of linear momentum;

mu₁ + m₂u₂ = mv₁ + m₂v₂

mu₁  +  0  = 0.45u₁m + m₂v₂

mu₁  = 0.45u₁m + m₂v₂ -------- equation (i)

Velocity for elastic collision in one dimension;

u₁ + v₁ = u₂ + v₂

u₁ + 0.45u₁ = 0 + v₂

1.45u₁ = v₂ (final velocity of the second ball)

Substitute in v₂ into equation (i)

mu₁  = 0.45u₁m + m₂(1.45u₁)

mu₁ = 0.45u₁m + 1.45m₂u₁

mu₁ - 0.45u₁m = 1.45m₂u₁

0.55mu₁ = 1.45m₂u₁

divide both sides by u₁

0.55m = 1.45m₂

m₂ = 0.55m / 1.45

m₂ = 0.379m

Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)

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Answer:

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Explanation:

The following situations apply: A, B, D.

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2.) As for B, the electric force on the electron depends on its charge as you can see from equation above.

3.) As for C, mass doesn't appear anyway in the equation for Force, thus, the mass doesn't matter.

4.) For D, yes, see equation for Force.

5.) For E, no

6.) For F, r in the equation for the force refers to the distance between both charges, not the radius of either charge. Thus, F doesn't matter.

7.) For G, no.

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Suppose (for this statement only), that q is moved from the origin but is still within both the surfaces. The flux through both
o-na [289]

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-True  - True  - true   - false -false - false

Explanation:

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2 years ago
he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee
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Answer:

a

 F =  -1.07 *10^{-8} \  N

b

F_r  =  1.07 *10^{-8} \  N

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

F =  \frac{k *  [Q_{Li}  ] * [Q_{O}  ]  }{ r^2}

Here k is known as the proportionality constant with value k = 2.31  *  10^ {-28} J \cdot m

substituting -2 for Q_{O} i.e the charge on oxygen , +1 for Q_{Li} i.e the charge on Lithium and [0.140 + 0.068 ] nm= 0.208 nm =  0.208*10^{-9} for r

So

F =  \frac{ 2.31  *  10^ {-28}*  +1   * -2   }{ ( 0.208*10^{-9} )^2   }

F =  -1.07 *10^{-8} \  N

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

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6 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

       

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3 years ago
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