Answer: -31.36 m/s
Explanation:
This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:
(1)
Where:
is the final velocity of the supply bag
is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)
is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)
is the time
Knowing this, let's solve (1):
(2)
Finally:
Note the negative sign is because the direction of the bag is downwards as well.
"The" (and any subsequent words) was ignored because we limit queries to 32 words.
Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
S=Vt
110=V(72)
110/72=V
V=1.527m/s
Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.