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4 years ago

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A 1530 kg car moving south at 12.1 m/s collides with a 2560 kg car moving north. The cars stick together and move as a unit afte

r the collision at a velocity of 5.48 m/s to the north. Find the velocity of the 2560 kg car before the collision. Assume that North is positive. Answer in units of m/s.

1 answer:

sveticcg [70]4 years ago

8 0

Answer:

16 m/s north

Explanation:

According to the law of momentum conservation, total momentum before and after must be the same

Total momentum after the collision as both cars stick together as 1 unit traveling at speed of 5.48m/s to the north

(m + M)V = (1530 + 2560)*5.48 = 22413 kgm/s

So the total momentum before must also be 22413, which is made of

$mv_1 + Mv_2 = 22413$

$1530*(-12.1) + 2560v_2 = 22413$

$2560v_2 = 22413 + 18513 = 40926$

$v_2 = 40926 / 2560 \approx 16 m/s$ north

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3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

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3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

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where $V_{1x}$ = velocity in the X - direction

t = Time taken

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Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

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Making time (t) subject of the formula in Equation 1

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substituting equation 3 into equation 2

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Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

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In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

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Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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3 years ago