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AlekseyPX
3 years ago
15

You move a 25 n object 4 meters. find the work you did

Physics
2 answers:
Alekssandra [29.7K]3 years ago
6 0
Hello there.

You move a 25 n object 4 meters. find the work you did<span>
100 N.m</span>
d1i1m1o1n [39]3 years ago
5 0
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:

Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
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inysia [295]

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

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If you weigh 400 N on the Earth, what would you weigh on the moon?
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14.87 pounds or 66.17 n

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What is the acceleration of a ball thrown vertically upward in the presence of Earth’s
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2 years ago
Please I need it now​
Nana76 [90]

1. A vector \vec x is a unit vector if its magnitude is 1. Given

\vec A = \dfrac{\sqrt3}2\,\vec\imath - \dfrac12\,\vec\jmath \\\\ \vec B = -\dfrac{\sqrt2}2\,\vec\imath - \dfrac{\sqrt2}2\,\vec\jmath \\\\ \vec C = \dfrac12\,\vec\imath - \dfrac{\sqrt3}2\,\vec\jmath - \dfrac12\,\vec k

\vec A and \vec B are unit vectors, while \vec C is not, since

\|\vec A\| = \sqrt{\left(\dfrac{\sqrt3}2\right)^2 + \left(-\dfrac12\right)^2} = 1 \\\\ \|\vec B\| = \sqrt{\left(-\dfrac{\sqrt2}2\right)^2 + \left(-\dfrac{\sqrt2}2\right)^2} = 1 \\\\ \|\vec C\| = \sqrt{\left(\dfrac12\right)^2+\left(-\dfrac{\sqrt3}2\right)^2+\left(-\dfrac12\right)^2} =\dfrac{\sqrt5}2 \neq 1

2. Given some vector \vec x, you can get the unit vector in the same direction as

If

\vec A = -12\,\vec\imath + 9\,\vec\jmath \\\\ \vec B = 15\,\vec\imath-20\,\vec\jmath

then the unit vectors in the direction of \vec A and \vec B, respectively, are

\dfrac{\vec A}{\|\vec A\|} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{\sqrt{(-12)^2+9^2}} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{15} = -\dfrac45\,\vec\imath + \dfrac35\,\vec\jmath \\\\ \dfrac{\vec B}{\|\vec B\|} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{\sqrt{15^2+(-20)^2}} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{25} = \dfrac35\,\vec\imath - \dfrac45\,\vec\jmath

3.

\vec A = \|\vec A\| \left(\cos(180^\circ-\theta)\,\vec\imath + \sin(180^\circ-\theta)\,\vec\jmath\right) \\\\ \vec A = 25 \left(-\sin(37^\circ)\,\vec\imath + \cos(37^\circ)\,\vec\jmath\right) \\\\ \vec A = -15\,\vec\imath + 20\,\vec\jmath

\vec B = \|\vec B\| \left(\cos(180^\circ + \alpha)\,\vec\imath + \sin(180^\circ+\alpha)\,\vec\jmath\right) \\\\ \vec B = 75 \left(-\sin(53^\circ)\,\vec\imath-\cos(53^\circ)\,\vec\jmath\right) \\\\ \vec B = -60\,\vec\imath - 45\,\vec\jmath

6 0
2 years ago
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