<span>a. The magnitude of the vector is doubled as well.
Let's say we have a 2-dimensional vector with components x and y.
It's magnitude lâ‚ is given by:
lâ‚ = âš(x² + y²)
If we double the components x and y, the new magnitude lâ‚‚ is:
lâ‚‚ = âš((2x)² + (2y²))
With a bit of algebra...
lâ‚‚ = âš(4x² + 4y²)
lâ‚‚ = âš4(x² + y²)
lâ‚‚ = 2âš(x² + y²)
We can write the new magnitude lâ‚‚ in terms of the old magnitude lâ‚.
lâ‚‚ = 2lâ‚
Therefore, the new magnitude is double the old one.
It should be clear that this relationship applies to 3D (and 1D) vectors as well.
b. The direction angle is unchanged.
The direction angle θ₠for a 2-dimensional vector is given by:
θ₠= arctan(y / x)
If we double both components, we get:
θ₂ = arctan(2y / 2x)
θ₂ = arctan(y / x)
θ₂ = θâ‚
The new direction angle is the same as the old one.</span>
Not a full question that can be answered sorry:/
KE= 0.5mv^2
165J= 0.5×(825kg+1kg)×v^2
165J = 413kg×v^2
v^2=165J÷413kg
v^2=0.39952 m^2/s^2 (5s.f.)
v= √0.39952= 0.632 m/s (3.s.f.)
