A bicycle of mass m requires 50 J of work to move from rest to a final speed v. If the same amount of work is performed during the same amount of time on a bicycle with a mass of 2m, what is the final speed of the second bicycle?

The final speed of the second bicycle is (v·√2)/2

Explanation:

The mass of the given bicycle = m

The amount of work required to move the bicycle from rest to speed v = 50 J

The final speed of the first bicycle = v

The mass of the second bicycle = 2m

Therefore, from conservation of energy, we have;

Work required by the first bicycle = Kinetic energy gained by the bicycle

The kinetic energy = 1/2·m·v²

∴ Energy required by the first bicycle = 50 J = 1/2·m·v²

Given that the same amount of work is performed on the second bicycle, we have;

Work performed on the second bicycle = 50 J = kinetic energy of second bicycle = 1/2·(2·m)·v₂²

Also, given that 50 J = 1/2·m·v², we have;

Work performed on the second bicycle = 50 J = 1/2·m·v²= 1/2·(2·m)·v₂²

1/2·m·v²= 1/2·(2·m)·v₂²

m·v² = 2·m·v₂²

v² = 2·v₂²

v₂ = √(v²/2) = v/√2 = (v·√2)/2

v₂ = (v·√2)/2

The final speed of the second bicycle = v₂ = (v·√2)/2.

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

the answer is b i think