Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both dogs sprint away in different directions while Daniel stands still. From Daniel's point of view, Newton runs due North at 3.903.90 m/s, but from Pauli's point of view, Newton appears to be moving at 1.501.50 m/s due East. What must Pauli's velocity relative to Daniel be for this to be true? Express your answer in terms of the xx ‑ and yy ‑components if North is the +y+y ‑direction and East is the +x+x ‑direction.

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

x-component = -1.50 m/s

y-component = 3.90 m/s

Explanation:

Relative velocity of a body A relative to another body B, Vab, is given as

Vab = Va - Vb

where

Va = Relative velocity of body A with respect to another third body or frame of reference C

Vb = Relative velocity of body B with respect to that same third body or frame of reference C.

So, relative velocity can be given further as

Vab = Vac - Vbc

Velocity of Newton relative to Daniel = Vnd = 3.90 m/s due north = (3.90ĵ) m/s in vector form.

Velocity of Newton relative to Pauli = Vnp = 1.50 m/s due East = (1.50î) m/s in vector form

What is Pauli's velocity relative to Daniel?

Vpd = Vp - Vd

(Pauli's velocity relative to Daniel) = (Pauli's velocity relative to Newton) - (Daniel's velocity relative to Newton)

Vpd = Vpn - Vdn

Vpn = -Vnp = -(1.50î) m/s

Vdn = -Vnd = -(3.90ĵ) m/s

Vpd = -1.50î - (-3.90ĵ)

Velocity of Pauli relative to Daniel = (-1.50ï + 3.90ĵ) m/s

Hope this Helps

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

measure their densities same density, same sub