Answer:
2 m/s^2
Explanation:
a = v^2/r
a = (10m/s)^2 / 50m
a = 2 m/s^2
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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.
A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters
Answer:
The first frequency of audible sound in the speed sound is
f = 662 Hz
Explanation:
vs = 344 m/s
x = 52 cm * 1 / 100m = 0.52m
The wave length is the distance between the peak and peak so
d = 2x
d = 2*0.52 m
d = 1.04 m
So the frequency in the speed velocity is
f = 1 / T
f = vs / x = 344 m/s / 0.52m
f ≅ 662 Hz
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
