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1 year ago

Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k

Physics

1 answer:

Sliva [168]1 year ago

5 0

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

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What is the mechanical advantage of a wedge that is 2 inches at its widest part and has a sloped side with a length of 10 inches

Delicious77 [7]

The correct answer would be 5 hope this helps.

7 0

3 years ago

Read 2 more answers

honor physics The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force. a. Wha

Ksivusya [100]

Answer:

(a) 4000 kgm/s.

(b) 80 N

Explanation:

(a) Change in momentum: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is Ns or kgm/s

Mathematically, Change in momentum is expressed as

ΔM = mΔv ..................................... Equation 1

Where ΔM = change in momentum, m = mass of the object, Δv = change in velocity = v₂ - v₁

Given: m = 200 kg, Δv = v₂ - v₁ = 25-5 = 20 m/s.

Substituting into equation 1

ΔM = 200(20)

ΔM = 4000 kgm/s.

Hence the change in momentum = 4000 kgm/s

(b)

Force: This can be defined as the ratio of the change in momentum of a body to the time required for the change.

F = ΔM/t.............................. Equation 2

Where F = force, ΔM = change in momentum, t = time.

Given: ΔM = 4000 kgm/s, t = 50 second.

Substituting into equation 2

F = 4000/50

F = 80 N.

Hence the force = 80 N

7 0

3 years ago

If Galileo drops a cannon ball from the 60 meter high) Leaning Tower of Pisa, how fast will it be moving when it hits the ground

viva [34]

Answer:

When the ball hits the ground, the velocity will be -34 m/s.

Explanation:

The height and velocity of the ball at any time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity. (-9.8 m/s² considering the upward direction as positive).

v = velocity at time "t".

If we place the origin of the frame of reference on the ground, when the ball hits the ground its height will be 0. Then using the equation of height, we can calculate the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

0 = 60 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

0 = 60 m - 4.9 m/s² · t²

-60 m / -4.9 m/s² = t²

t = 3.5 s

Now, with this time, we can calculate the velocity of the ball when it reaches the ground:

v = v0 + g · t

v = 0 m/s - 9.8 m/s² · 3.5 s

v = -34 m/s

When the ball hits the ground, the velocity will be -34 m/s.

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3 years ago

If u have 2500Watts,240Volts . its resistance

Virty [35]

Answer:

2500/240 will give the resistance to be 10.42

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3 years ago

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

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3 years ago