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3 years ago

In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.

Physics

2 answers:

Free_Kalibri [48]3 years ago

6 0

In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.

TRUE.

Bess [88]3 years ago

4 0

Answer: True

Explanation:

A chemical reaction in which the oxidation number of an atom, ion or molecule changes due transfer of electrons between two reactants is known as oxidation-reduction reaction or redox reaction. In Redox reaction, the oxidising agent brings oxidation by gaining electrons and reducing it self.

Example:

Zn + Cu²⁺ → Zn²⁺ + Cu

Here, Cu²⁺ is an oxidizing agent which got reduced to Cu .

Thus, the given statement is true.

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A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a

Zolol [24]

Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

$E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L$

For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

$\theta = \frac{A}{L^{2}}$ , measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is $4\theta$ and the total energy of the pendulum is:

$E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L$

The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

3 0

3 years ago

Think of three questions that will help you clarify your understanding of how of the factors that have contributed to the rise i

fenix001 [56]

Answer:

the glucose in air will solidify and it will stop the process.

it will cause hot air

6 0

2 years ago

Read 2 more answers

Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25

dsp73

Here,

height at failure, h1 = 525 m,
upward acceleration, a = 2.25 m/s^2,
velocity = v m/s,

SO, 

v^2 = 2*a*h = 2*2.25*525 = 2362.5 
Now, acceleration, g = 9.8 m/s^2,

SO, 

heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters 
Hence,

a) 
Total height = 525+120.54 = 645.54 meters

b)
time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec

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4 0

3 years ago

Read 2 more answers

If a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please

kherson [118]

Answer:

a = 3.125 [m/s^2]

Explanation:

In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.

$90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} \\= 25 \frac{m}{s}$

$v_{f} =v_{i} + (a*t)$

where:

Vf = final velocity = 25 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 8 [s]

The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.

25 = 0 + a*8

a = 3.125 [m/s^2]

3 0

2 years ago

What is the acceleration of a body moving with uniform velocity?

AfilCa [17]

Acceleration is the rate of change of velocity, a body moving with uniform velocity does not possess acceleration at all i.e. acceleration is zero

3 0

2 years ago