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solniwko [45]
2 years ago
9

If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B​​​

Physics
1 answer:
postnew [5]2 years ago
8 0

Let <em>B</em> = <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em>. Then the cross product of <em>A</em> = <em>i</em> + 2<em>j</em> - <em>k</em> with <em>B</em> is

<em>A</em> × <em>B</em> = ( <em>i</em> + 2<em>j</em> - <em>k </em>) × ( <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em> )

<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

<em>A</em> × <em>B</em> = 0 - 2<em>a</em> <em>k </em>- <em>a</em> <em>j</em>

… … … + <em>b</em> <em>k</em> + 0 + <em>b</em> <em>i</em>

… … … - <em>c</em> <em>j</em> + 2<em>c</em> <em>i</em> - 0

<em>A</em> × <em>B</em> = (<em>b</em> + 2<em>c</em>) <em>i</em> + (-<em>a</em> - <em>c</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

So we have

3 <em>i</em> - <em>j</em> + 5 <em>k </em>= (<em>b</em> + 2<em>c</em>) <em>i</em> + (<em>c</em> - <em>a</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

which gives us the system of equations,

{ <em>b</em> + 2<em>c</em> = 3

{ -<em>a</em> - <em>c</em> = -1

{ -2<em>a</em> + <em>b</em> = 5

Solve for <em>a</em>, <em>b</em>, and <em>c</em>.

• Eliminate <em>c</em> from the first two equations:

(<em>b</em> + 2<em>c</em>) + 2 (-<em>a</em> - <em>c</em>) = 3 + 2 (-1)

-2<em>a</em> + <em>b</em> = 1

But -2<em>a</em> + <em>b</em> = 5, and 5 ≠ 1, so there is no such vector <em>B</em> that satisfies the cross product!

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At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.
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3 years ago
Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i
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L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

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I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

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3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

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