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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Answer:
a) = 258352.5J
b) = 23.63 m/s
c) = 1.8m
Explanation:
Data;
Mass = 925kg
Distance (s) = 28.5m
Force constant (k) = 8.0*10⁴ N/m
g = 9.8 m/s²
a) = work = force * distance
But force = mass * acceleration
Force = 925 * 9.8 = 9065N
Work = F * s = 9065 * 28.5 = 258352.5J
b) acceleration (a) = (v² - u²) / 2s
a = v² / 2s
v² = a * 2s
v² = 9.8 * (2 * 28.5)
v² = 9.8 * 57
v² = 558.6
v = √(558.6)
V = 23.63 m/s
C). The work stops when the work done to raise the spring equals the work done to stop it by the spring
W = ½kx²
258352.5 = ½ * 8.0*10⁴ * x²
(2 * 258352.5) = 8.0*10⁴x²
516705 = 8.0*10⁴x²
X² = 516705 / 8.0*10⁴
X² = 6.46
X = √(6.46)
X = 2.54m
The compression was about 2.54m
Answer:
d. equal to one-fourth the acceleration at the surface of the asteroid.
Explanation:
The explanation is attached as a picture with this answer
Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.
as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.
As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.
It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
There are none on the list you included with your question.