The second stone hits the ground exactly one second after the first.
The distance traveled by each stone down the cliff is calculated using second kinematic equation;

where;
- <em>t is the time of motion </em>
- <em />
<em> is the initial vertical velocity of the stone = 0</em>

The time taken by the first stone to hit the ground is calculated as;

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as


Thus, we can conclude that the second stone hits the ground exactly one second after the first.
"<em>Your question is not complete, it seems be missing the following information;"</em>
A. The second stone hits the ground exactly one second after the first.
B. The second stone hits the ground less than one second after the first
C. The second stone hits the ground more than one second after the first.
D. The second stone hits the ground at the same time as the first.
Learn more here:brainly.com/question/16793944
In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.
The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.
A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.
It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:
-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .
Answer: C. Meyer did not leave gaps while Mendeleev did.
Explanation: Mendeleev's table was table different from Meyer's table because Meyer did not leave gaps while Mendeleev did.
Answer:
1 ohm
Explanation:
First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:
1 / Eq = (1 / r₁) + (1 / r₂)
The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:
Eq = r₁ + r₂
Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:
1/E₁ = 1/2 + 1/2
1/E₁ = 1
E₁ = 1Ω
1/E₂ = 1/2 + 1/2
1/E₂ = 1
E₂ = 1Ω
This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:
E₃ = E₁ + E₂ = 1 + 1 = 2Ω
The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:
1/Eq = 1/E₃ + 1/2
1/Eq = 1/2 + 1/2
1/Eq = 1
Eq = 1Ω