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BabaBlast [244]
3 years ago
10

A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos

itive x-axis, 5 m, then goes back, 6 m. What is his total displacement from his original position? Displacement is a vector quantity whilst, distance is a scalar quantity.
Physics
1 answer:
Andreas93 [3]3 years ago
3 0

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

x" = -1 m

Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

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Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2  = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

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3 years ago
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft
Marina CMI [18]

Explanation:

The equation of motion of an object is given by :

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We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

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On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

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tankabanditka [31]
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Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
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If the moon were twice as far from years as it is now the following would be true
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The question is incomplete.

The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.

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