All categories

2 years ago

What occurs at the end of a negative cell if an electric cell is placed in a closed circuit?

1 answer:

4 0

1. Current means flow of electrons and electrons are negative charged and so are attracted to the positive end of the battery and repelled by the negative end.

2. When battery is connected in a circuit that lets the electron flow through it, they flow from negative to positive.

3. When negative terminal of cell is connected to other negative terminal of the cell in a particular circuit then, current will not flow in circuit as electrons cannot flow from negative to negative terminal.

You might be interested in

Help pleasseeee URGENT

Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

When a collision occurs, the velocities change to v' and the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

In a system of two masses, the law of conservation of linear momentum is simplified to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

$\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}$

$\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}$

$\displaystyle v'_1=\frac{0.34}{0.16}$

$v'_1=2.125\ m/s$

The speed of the 8-ball is 2.125 m/s after the collision.

8 0

3 years ago

Can someone pls help me with this? Its due in 25 minutes

olga_2 [115]

Answer:

Ty for the coins

Explanation:

3 0

2 years ago

A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axi

Vladimir [108]

Explanation:

It is given that,

Force applied to object, $F=(2i+3j)\ N$

Position, $r=(4i+5j)\ m$

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

$\tau=F\times r$

$\tau=(2i+3j) \times (4i+5j)$

$\tau=\begin{pmatrix}0&0&-2\end{pmatrix}$

$\tau=(-2k)\ N-m$

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

$|\tau|=\sqrt{(-2)^2}$

$|\tau|=2\ N-m$

Hence, this is the required solution.

5 0

3 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

2 years ago

Predict the energy transformation that occurs when carson sands a piece of wood with sandpaper.

tigry1 [53]

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

6 0

2 years ago

What occurs at the end of a negative cell if an electric cell is placed in a closed circuit?​

What occurs at the end of a negative cell if an electric cell is placed in a closed circuit?