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3 years ago

How are the current moving in the simulation of battery

1 answer:

6 0

Answer:When the battery is connected to a circuit, electrons produced by the chemical reaction at the anode flow through the circuit to the cathode. ... Batteries put out direct current, as opposed to alternating current, which is what comes out of a wall socket. With direct current, the charge flows only in one direction.

Explanation:

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A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a sp

cluponka [151]

Answer:

(a) Approximately $0.335\; \rm m$ .

(b) Approximately $1.86\; \rm m\cdot s^{-1}$ .

(c) Approximately $0.707\; \rm m$ .

(d) Approximately $0.228\; \rm m$ .

Explanation:

$v_i$ denotes the velocity of the object in the first diagram right before it came into contact with the spring.
Let $m$ denote the mass of the block.
Let $\mu$ denote the constant of kinetic friction between the object and the surface.
Let $g$ denote the constant of gravitational acceleration.
Let $k$ denote the spring constant of this spring.

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy $\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2$ .

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of $D$ and compressed the spring by the same distance.

Energy lost to friction: $\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D$ .
Elastic potential energy that the spring has gained: $\displaystyle \frac{1}{2}\,k\, D^2$ .

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

$\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

Assume that $g = 9.81\; \rm m \cdot s^{-2}$ . In the equation above, all symbols other than $D$ have known values:

$m =1.10\; \rm kg$ .
$v_i = 2.60\; \rm m \cdot s^{-1}$ .
$\mu = 0.250$ .
$g = 9.81\; \rm m \cdot s^{-2}$ .
$k = 50.0\; \rm N \cdot m^{-1}$ .

Substitute in the known values to obtain an equation for $D$ (where the unit of $D\!$ is $m$ .)

$3.178 = 2.69775\, D + 25\, D^2$ .

$2.69775\, D + 25\, D^2 + 3.178 = 0$ .

Simplify and solve for $D$ . Note that $D > 0$ because the energy lost to friction should be greater than zero.

$D \approx 0.335\; \rm m$ .

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

$\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J$ .

As the object moves to the left, part of that energy will be lost to friction:

$(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J$ .

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

$2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J$ .

Calculate the velocity corresponding to that kinetic energy:

$\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}$ .

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ( $1.91\; \rm J$ ) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is $\mu \cdot m \cdot g$ .

$\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m$ .

Similar to (a), solving (d) involves another quadratic equation about $D$ .

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) $1.91\; \rm J$ .

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

$\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

$25\, D^2 + 2.69775\, D - 1.90811\approx 0$ .

Again, $D > 0$ because the energy lost to friction is greater than zero.

$D \approx 0.228\; \rm m$ .

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3 years ago

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