A0.64 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

... If nothing is moving, then nothing is accelerating.

... If nothing is accelerating, then the net force on the frog is zero,
and we know the forces on the frog's posterior are balanced.

... So the normal force of the log acting upward on the frog is
identical to the weight of the frog acting downward on the log.

... The frog's weight is (mass) x (gravity)

                                 =  (0.64 kg) x (9.8 m/s²)

                                 =    6.27 kg-m/s²

                                 =     6.27 Newtons

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.