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Amanda [17]
3 years ago
5

Find a unit vector normal to the plane containing Bold u equals 2 Bold i minus Bold j minus 3 Bold k and Bold v equals negative

3 Bold i plus Bold j minus 2 Bold k.
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Explanation:

It is given that,

\vec{u}=2i-j-3k

\vec{v}=-3i+j-2k

Taking the cross product of v and v such that,

\vec{w}=u\times v

\vec{w}=(2i-j-3k)\times (-3i+j-2k)

\vec{w}=5i+13j-k

|w|=\sqrt{5^2+13^2(-1)^2}

|w| = 13.92

Let \hat{w} is the unit vector normal to the plane containing u and v. So,

\hat{w}=\dfrac{\vec{w}}{|w|}

\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}

Hence, this is the required solution.

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Name nine elements that have been around so long that we don't even know when they
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A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

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pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

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g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

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the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

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