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4 years ago

8

What do you think the value will be for the car's distance

1 answer:

lions [1.4K]4 years ago

8 0

Answer:

student A

Explanation:

It's going the same value

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3 years ago

Answers are ONLY allowed to be ionic, polar covalent, and nonpolar covalent.

Luba_88 [7]

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2 years ago

He formula p=m x v to answer th<br>2,000 kg car moving at 30 m/s

aivan3 [116]

The momentum of the car is 60,000 kg m/s

Explanation:

The momentum of an object is a vector quantity, whose magnitude is given by

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where

m is the mass of the object

v is its velocity

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m = 2,000 kg

v = 30 m/s

Therefore, the magnitude of its momentum is

$p=(2000)(30)=60,000 kg m/s$

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3 years ago

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Alchen [17]

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3 0

3 years ago

Read 2 more answers

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago