A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What
is its potential energy (PE) when it
lands?

Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

Explanation:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its potential energy (PE) when it lands

The potential energy PE, relative to the ground, will be zero, because the lemming is at the ground level.

HOWEVER, a much better question would be:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its kinetic energy (KE) when it lands?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, u = 0, a = 9.81m/s^2, s = 5.36m

So we find v using equation (2)

v^2 = u^2 + 2as

v^2 = 0 + 2(9.81)(5.36) = 105.1632

So the kinetic energy resulting from the vertical drop is Ekv = ½mv^2

Ekv = ½(0.078)(105.16) = 4.101J

BUT we need to add in the kinetic energy resulting from the horizontal velocity, which did not change during the vertical drop.

Ekh = ½(0.078)(4.84^2) = 0.914J

So the total kinetic energy is Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

answer; the correct answer is technician a; ///100%good luck

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)