All categories

3 years ago

What role does induction play when lightning strikes earth?

Physics

1 answer:

Dominik [7]3 years ago

7 0

The role lightning plays in earth is when the earth is charged with positive protons, the lightning is the electron.

You might be interested in

A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Wr

Monica [59]

Answer : The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

The given redox reaction occurs between the magnesium and cadmium.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$

4 0

3 years ago

Read 2 more answers

PLS HELP!!!! If u help I will mark brainiest thx

tatyana61 [14]

Answer:

Semiconductor. sorry i'm late

Explanation:

8 0

4 years ago

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

3 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0

3 years ago

Can someone help me pls

inna [77]

Answer:

Explanation:

8 0

3 years ago