Where does reduction occur in an electrolytic cell

Sladkaya [172]

Answer:

a change of position in relation to another object.

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

A reference point refers to a location or physical object from which the motion (movement) of another physical object or body can be determined.

Mathematically, the motion of an object is described in terms of acceleration, time, distance, speed, velocity, displacement, position, etc.

Hence, motion is a change of position in relation to another object.

7 0

2 years ago

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 90.0 m above the

romanna [79]

Answer:

d = 771.3m

Explanation:

Let's first calculate the time of flight:

$Yf = Yo+Voy*t-g*t^2/2$ where Voy=0. Solving for t:

$t = \sqrt{2*90/9.8} =4.285s$

Now we calculate the horizontal displacement, wihch is the distance from the target to drop the package:

Xf = d = Vox*t

d = 180*4.285 = 771.3m

6 0

3 years ago

Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const

IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

$\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}$

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

$\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}$

Therefore the speed is 1.026 m/s.

5 0

1 year ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

Why do electric fish need to force electric charges to move?

Anastasy [175]

The electric eel generates large electric currents by way of a highly specialized nervous system that has the capacity to synchronize the activity of disc-shaped, electricity-producing cells packed into a specialized electric organ. Hope this helps!! Good luck

3 0

3 years ago