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Alex Ar [27]
3 years ago
12

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

astic, the ball bounces back with the same speed v going upward. Which of the following statements about the bounce are correct? (There could be more than one correct choice.) 1. The balrs mpmentum was conserved during the bounce. 2. The ball ha the same momentum just before and just after the bounce. 3. The magnitude of the ball's momentum was the same just before and just after the bounce. 4. All of the above 5. None of the above
Physics
1 answer:
artcher [175]3 years ago
7 0

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

                     p₀ = m v

The final moment

                     p_{f} = m (-v)

 

                  p₀ = - p_{f}

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

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True

Explanation:

Golden Rice is an example of genetically modified organism.

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3 years ago
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The force of attraction between a ball is F=.........×10^-¹¹
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Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

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2 years ago
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Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
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Answer:

1115560000 J

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3 years ago
A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge
Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance C_1 with energy of 7.54 J, is connected in parallel with another capacitor C_2 , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

U_O=q_0^2/2C_1=7.54 J\\

The energy stored in the electric field is the sum of the energies of the two capacitors:

U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2

since the charge equally distributed,  q_1 = q_2 = q_o/2. and since they are connected in parallel the potential difference on both of them is the same :

V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\

hence,

U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J

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