Duracell batteries are an example of an electrochemical cell that is powered between the reaction of Magnesium and Zinc, occurring in basic conditions (alkaline battery). This type of reaction has a precise output of 1.5 volts, and looks like this:
Zn + 2MnO2 ➡️ ZnO + Mn2O3
It’s not rechargeable.
Golf Cart Batteries are an example of an electrochemical cell that is powered by the reaction between Lead and Sulfuric Acid (Lead-Acid battery). This type of reaction occurs on larger scales than an alkaline battery, and thus can generate a variety of powers depending on how many instruments are present within the battery. The reaction looks like this:
PbO2 + Pb + 2H2SO4 ➡️ 2PbSO4 + H2O
This is a rechargeable cell, but is rather prone to discharging by the environment and surroundings of the battery.
Answer:
At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,
The velocity will then increase to the midpoint of the motion.
m g h = 1/2 m v^2 where h is the vertical distance thru which the pendulum travels
The maximum height reached by the ball is 99.2 m
Explanation:
When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (
towards the ground). Therefore, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
In this problem, we have:
u = 44.1 m/s is the initial vertical velocity of the ball
v = 0 is the final velocity when the ball reaches the maximum height
s is the maximum height
is the acceleration of gravity (downward, so negative)
Solving for s, we find the maximum height reached by the ball:
Learn more about free fall:
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Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Answer:
The force required to push to stop the car is 288.67 N
Explanation:
Given that
Mass of the car, m = 1000 kg
Initial speed of the car, u = 1 m/s
The car and push on the hood at an angle of 30° below horizontal, 
Distance, d = 2 m
Let F is the force must you push to stop the car.
According work energy theorem theorem, the work done is equal to the change in kinetic energy as :
The force required to push to stop the car is 288.67 N
