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2 years ago

50 POINTS!!!!!

Physics

2 answers:

FinnZ [79.3K]2 years ago

8 0

Answer:

The answer is C.move forward rapidly until stopped.

Explanation:

50 miles an hour is a very high speed. The momentum transfer to the car at stop light will move the car forward rapidly until its brake is able to stop it.

n200080 [17]2 years ago

4 0

Answer:

Explanation:

law of conservation of momentum says the car that u hit will move forward rapidly. then it wuld stop. ans is C

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Rodney is trying out one of Santa's new games that consists of three pieces that blow apart if the wrong key is inserted into th

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Answer:

<em>The third piece moves at 6.36 m/s at an angle of 65° below the horizon</em>

Explanation:

Linear Momentum

It's a physical magnitude that measures the product of the velocity by the mass of a moving object. In a system where no external forces are acting, the total momentum remains unchanged regardless of the interactions between the objects in the system.

If the velocity of an object of mass m is $\vec v$ , the linear momentum is computed by

$\displaystyle \vec{P}=m.\vec{v}$

The momentum of the board before the explosion is

$\displaystyle \vec{P}_{t1}=m_t\ \vec{v}_o$

Since the board was initially at rest

$\displaystyle \vec{P}_{t1}=$

After the explosion, 3 pieces are propelled in different directions and velocities, and the total momentum is

$\displaystyle \vec{P}_{t2}=m_1\ \vec{v}_1\ +\ m_2\ \vec{v}_2+m_3\ \vec{v}_3$

The first piece of 2 kg moves at 10 m/s in a 60° direction

$\displaystyle \vec{v}_1=(10\ m/s,60^o)$

We find the components of that velocity

$\displaystyle \vec{v}_1=$

$\displaystyle \vec{v}_1=m/s$

The second piece of 1.2 kg goes at 15 m/s in a 180° direction

$\displaystyle \vec{v}_2=(15,180^o)$

Its components are computed

$\displaystyle \vec{v}_2=(15\ cos180^o,15\ sin180^o)$

$\displaystyle \vec{v}_2=(-15,0)\ m/s$

The total momentum becomes

$\displaystyle P_{t2}=2+1.2+m_3\ \vec{v}_3$

Operating

$\displaystyle P_{t2}=++m_3\ \vec{v}_3$

Knowing the total momentum equals the initial momentum

$\displaystyle P_{t2}=+m_3\ \vec{v}_3=0$

Rearranging

$\displaystyle m_3\ \vec{v}_3=$

Calculating

$\displaystyle m_3\ \vec{v}_3=$

This is the momentum of the third piece

From the above equation, we solve for $\vec v_3$ :

$\displaystyle \vec{v}_3=\frac{1}{3}$

$\displaystyle \vec{v}_3=m/s$

The magnitude of the velocity is

$\displaystyle \vec{v}_3|=\sqrt{2.67^2+(-5.77)^2}=6.36$

And the angle is

$\displaystyle tan\theta =\frac{-5.77}{2.67}=-2.161$

$\displaystyle \theta =-65.17^o$

The third piece moves at 6.36 m/s at an angle of 65° below the horizon

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