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ICE Princess25 [194]
3 years ago
6

Two ice skaters, paula and ricardo, push off from each other. ricardo weighs more than paula. part a which skater, if either, ha

s the greater momentum after the push-off? which skater, if either, has the greater momentum after the push-off? paula has the greater momentum. ricardo has the greater momentum. both skaters have momentums of equal magnitude. submitmy answersgive up part b which skater, if either, has the greater speed after the push-off? which skater, if either, has the greater speed after the push-off? paula has the greater speed. ricardo has the greater speed. both skaters have equal speeds.
Physics
1 answer:
ahrayia [7]3 years ago
7 0

Apply the law of conservation of momentum for this situation. The law states that the momentum of a system is constant (in absence of external forces acting on it).

The 'system' in this case are the two skaters. There is no external force on the skaters. Suppose the skaters are initially standing still. The momentum in the system is 0. This value will need to remain constant, even after the mutual push (which is a set of forces from <em>inside</em> the system). So we know that

(total momentum before) = (total momentum after)

Indexing the masses and velocities by the first letter of the skaters' names:

0 = m_P\vec v_P+m_R\vec v_R\\m_P\vec v_P = m_R(-\vec v_R)

From the last row, you can see that the skaters will have momentum of same magnitude but opposite direction, after the push off. That answers the first question: neither will have a greater momentum (both will have one of same magnitude).

Since Ricardo is heavier, from the above equality it follows that

m_R>m_P\implies|\vec v_R|

In words, Paula has the greater speed, after the push-off.

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A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb
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Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

n_1 =\dfrac{46 \times 1}{0.0821\times 298}

  n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

n_2 =\dfrac{12 \times 1}{0.0821\times 298}

  n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

P_1=\dfrac{n_1 R T}{V}

P_1=\dfrac{1.89\times 0.0821\times 298}{5}

     P₁ = 9.25 atm

Partial pressure of oxygen

P_2=\dfrac{n_2 R T}{V}

P_2=\dfrac{0.49\times 0.0821\times 298}{5}

    P₂ = 2.39 atm

total pressure

    P = P₁ + P₂

    P = 9.25 + 2.39

    P = 11.64 atm

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A mountain climber at the peak ha ___________ energy.
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Its Kinetic, hope this helps you
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What kind of device can you use to to separate visible light into its different colors
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A prism will separate white light into a rainbow of light
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Titanium is a metal used to make golf clubs. A rectangular bar of this metal measuring 1.96 cm x 2.19 cm x 2.63 cm was found to
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4.012\frac{gr}{cm^3}

Explanation

the density of an object is given by:

\text{Density(d)}=\frac{mass(m)}{\text{volume(v)}}

Step 1

find the volume of the bar

a)find the volume of the rectangular bar.

the volume of a rectangular prism is given by:

\text{Volume}=\text{ length}\cdot widht\cdot depth

replace

\begin{gathered} \text{Volume}=(\text{ 2.63}\cdot2.19\cdot1.96)(cm^3) \\ \text{Volume}=11.289012(cm^3) \end{gathered}

Step 2

now,

Let

\begin{gathered} \text{Volume}=11.289012(cm^3) \\ \text{mass}=\text{ 45.3 gr} \end{gathered}

replace in the formula

\begin{gathered} \text{Density(d)}=\frac{mass(m)}{\text{volume(v)}} \\ d=\frac{45.3\text{ gr}}{11.289012(cm^3)} \\ d=4.012\frac{gr}{cm^3} \end{gathered}

therefore, the answer is

4.012\frac{gr}{cm^3}

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1 year ago
An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform
frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

E_p = E_e

mgh = kx^2/2

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

80*9.81*2.74 = k0.24^2/2

2150.352 = 0.0288k

k = 74665 N/m

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

E_p = E_{e2} + E_k + E_{p2}

mgh = kx_2^2 + mv^2/2 + mgh_2

2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12

2150.352 = 537.588 + 40v^2 + 94.176

40v^2 = 1518.588

v^2 = 37.9647

v = \sqrt{37.9647} = 6.16 m/s

When he steps gently, then his gravity force would equal to his spring force

mg = kx_3

x_3 = mg/k = 80*9.81/74665 = 0.0105m

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3 years ago
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