The characteristics of the scalar product allows to find the angle between the two vectors is:
The scalar product is the product between two vectors whose result is a scalar.
A . B = |A| |B| cos θ
Where A and B are the vectors, |A| and |B| are the modules of the vectors and θ at the angle between them.
The vector is given in Cartesian coordinates and the unit vectors in these coordinates are perpendicular.
i.i = j.j = 1
i.j = 0
A . B = (4 i - 4j). * -5 i + 7j)
A . B = - 4 5 - 4 7
A. B = -48
We look for the modulus of each vector.
|A| =
|A| =
|A| = 4 √2
|B| =
|B| = 8.60
We substitute.
-48 = 4√2 8.60 cos θ
-48 = 48.66 cos θ
θ = cos⁻¹
θ = 170º
In conclusion using the dot product we can find the angle between the two vectors is:
Learn more about the scalar product here: brainly.com/question/1550649
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Answer:
V₁ = 5.6 m/s
V₂ = 7.2 m/s
V₃ = 8.8 m/s
Explanation:
Average velocity: Average velocity can be defined as the ratio of the total displacement to the total time taken. The S.I unit of Average velocity is m/s.
For the first 2 s,
V₁ = Δd₁/t
Where V₁ = Average velocity for the first 2 s
Where Δd₁= distance, t = time
Δd₁ = 25.6-14.4 = 11.2 m t = 2 s
V₁ = 11.2/2
V₁ = 5.6 m/s
For the second 2 s,
V₂ =Δd₂/t
Where V₂ = average velocity for the second 2 s.
Δd₂= 40-25.6 = 14.4 m, t= 2 s
V₂ = 14.4/2
V₂ = 7.2 m/s
For the last 2 seconds,
V₃ =Δd₃/t
Where V₃ = average velocity for the last 2 s
where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s
V₃ = 17.6/2
V₃ = 8.8 m/s.
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.