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A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

Nataly [62]

Answer: $v_1=28.4 cm/s$

Explanation:

Given

mass of marble $m_1=27 gm$

velocity of marble $u_1=56.8 cm/s \approx 0.568 m/s$

mass of second marble $m_2=13.5 gm$

Velocity of second marble $u_2=14.2 cm/s \approx 0.142 m/s$

After collision 13.5 gm marble moves to the right at i.e. $v_2=71 cm/s$

Conserving momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71$

$1533.6+191.7=27\cdot v_1+958.5$

$27\cdot v_1=766.8$

$v_1=\frac{766.8}{27}=28.4 cm/s$

7 0

2 years ago

Does a comets tail always trail along behind it in its orbit?

Marat540 [252]

No, it only does when entering an atmosphere

4 0

2 years ago

Which statement about speed and/or velocity is true?

maw [93]

Answer: The right Answer is Velocity has both speed and direction.

Explanation:

i took the test

8 0

2 years ago

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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

2 years ago

The law of inertia to both moving and no moving objects. True or false

Zepler [3.9K]

True, the law of inertia effects both moving and non-moving objects.

3 0

2 years ago

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