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Ludmilka [50]
3 years ago
9

A 1.2-kg ball drops vertically onto the floor, hitting with a speed of 25 m/s. Consider the impulse during this collision. Would

the magnitude of the impulse be greater: (i) if the ball rebounded with a speed of 10 m/s (the ball was made of rubber), or (ii) if the ball stuck to the floor (the ball was made of clay)? Support your answer with a calculation.
Physics
1 answer:
Ksju [112]3 years ago
7 0

Answer:

3kg

Explanation:

impulse = MV

then

m1v1=m2v2

when the values are subtitude

then

m2=1.2*25/10

m2=<em>3</em><em>0</em><em>k</em><em>g</em><em>/</em><em>/</em>

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An elevator car has a mass of 750 kg, and its three passengers have a combined mass of 135 kg. If the elevator and its passenger
joja [24]

Answer:

The change in gravitational potential energy is -1.80x10⁵ J.

Explanation:

The change in gravitational potential energy is given by:

\Delta E_{p} = E_{p_{f}} - E_{p_{i}}

\Delta E_{p} = mgh_{f} - mgh_{i}

Where:

"i" is for final and "f" for final

m: is the mass

g: is the gravity = 9.81 m/s²

h: is the height

For the car and the passengers we have:

\Delta E_{p} = m_{T}g(h_{f} - h_{i}) = (750 kg + 135 kg)9.81 m/s^{2}(0 - 20.7 m) = -1.80 \cdot 10^{5} J      

The minus sign is because when the elevator car and the passengers are up they have a bigger gravitational potential energy than when they are in the ground.

Therefore, the change in gravitational potential energy is -1.80x10⁵ J.

I hope it helps you!                                                

3 0
3 years ago
What is the only possible value of ml for an electron in an s orbital?
Archy [21]

Answer:

  • zero

Explanation:

m_l     is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

  • n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

  • l : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

       This quantum number is related to the type (or shape) of the orbital:

        For s orbitals l=0

        For p orbitals l=1

        For d orbitals l=2

         For f orbitals l=3

In this case, it is an s orbital, so we have l=0.

  • m_l , the third quantum number can have integer values  {from-l}   to    {+l}

       Since, for the s orbitals  l=0 , the only possible value for {m_l} is zero.

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ANSWER:

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