Lucas programs his quadcopter to fly with a velocity (speed and direction) vector \vec{p} = 4\hat i + 7\hat j p =4 i ^ +7 j ^ p, with, vector, on top, equals, 4, i, with, hat, on top, plus, 7, j, with, hat, on top, which should send it straight at its target (in the absence of wind). To his dismay, his quadcopter actually moves with a velocity vector \vec{a} = 2\hat i + 8\hat j a =2 i ^ +8 j ^ a, with, vector, on top, equals, 2, i, with, hat, on top, plus, 8, j, with, hat, on top. (Speeds are given in meters per second, \text{m}/\text{s}m/sstart text, m, end text, slash, start text, s, end text.) What is the speed of the wind?

Velocity of the wind relative to the ground = (2î - ĵ) m/s

Speed of the wind = √5 m/s = 2.236 m/s

Step-by-step explanation:

Relative velocity of a body A with respect to B is given as the velocity of body A with respect to a reference frame C minus the velocity of body B with respect to a reference frame C.

That is

Vab = Vac - Vbc

Let the quadcopter be body A

The wind be body B

And the earth be the reference frame C.

Vac = velocity of the quadcopter relative to the ground = (4î + 7ĵ) m/s

Vab = velocity of the quadcopter with respect to the wind = (2î + 8ĵ) m/s

Vbc = velocity of the wind relative to the ground = ?

Vab = Vac - Vbc

(2î + 8ĵ) = (4î + 7ĵ) - Vbc

Vbc = (4î + 7ĵ) - (2î + 8ĵ)

Vbc = (2î - ĵ) m/s

The speed of the wind is the magnitude of its velocity

Magnitude of Vbc = √[2² + (-1)²] = √5 = 2.236 m/s

Hope this Helps!!!

4sqrt5

step-by-step explanation:

the base has dimensions of 8 x 4, so the diagonal is 64 + 16 square rooted or 4sqrt5.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.