A) The kinetic energy of an object is given by:

where m is the mass of the object, and v its speed. For the lion in our problem, m=45 kg and v=14.2 m/s, so its kinetic energy is

b) the increase in gravitational potential energy of the lion is given by:

where g is the gravitational acceleration, and

is the increase in altitude of the lion. In this problem,

, so the increase in gravitational potential energy is

c) When the fox reaches the top of the tree, its gravitational potential energy is

As it jumps, its kinetic energy is

So the total mechanical energy of the fox as it jumps is
Answer:
Because the zinc is reluctant
Explanation:
A leclanche cell contains a conducting solution (electrolyte) of ammonium chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide (oxidizer), and an anode (negative terminal) of zinc (reductant).
As the Zn2+ ions move away from the anode, leaving their electrons on its surface,
Zn → Zn2+ + 2e−
the anode becomes more negatively charged than the cathode. When the cell is connected to an external electrical circuit, the excess electrons on the zinc anode flow through the circuit to the carbon rod, the movement of electrons forming an electric current.
Answer:
80 meters high
Explanation:
The velocity of the balloon would be g*t (I won't calculate, but will us this later)
We know that the kinetic energy at the bottom equals the potential at the top.
KE = PE
1/2 * m * v^2 = m * g * h
1/2 * m * (g * t)^2 = m * g * h (substitution)
1/2 * m * g^2 * t^2 = m * g * h
1/2 * g * t^2 = h (simplification by dividing the commons between both sides)
h = 1/2 * 9.81 * 4^2
h = 78.48 m (roughly 80 m)
The answer of this is C!!!
Answer:
Part a)

Part b)

Part C)

Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as



now we will have




Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have


Part C)
By momentum conservation we will have



Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.