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4 years ago

The elements required for a fire to start are

Physics

2 answers:

denis-greek [22]4 years ago

8 0

Heat, Fuel, and Oxidizing agent

adelina 88 [10]4 years ago

4 0

The elements required for a fire to start are oxidizing agent, heat, and fuel.
The oxidizing agent is usually oxygen - when there is a lot of air, of oxygen, a fire can broke out more easily if the conditions are right - if there is no oxygen, the fire won't start - we put out a fire by depriving it of oxygen. Next, you need some heat and fuel to start a fire, which can all be seen in this triangle.

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Which form of energy is associated with an object’s position?

dybincka [34]

Answer:

potential

Explanation:

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3 years ago

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What must be the distance (in meters) between a point charge q1 = 16 μC and a point charge q2 = 32 μC for the electrostatic forc

adoni [48]

Answer:

d = 0.71 meters

Explanation:

It is given that,

Charge 1, $q_1=16\ \mu C=16\times 10^{-6}\ C$

Charge 2, $q_2=32\ \mu C=32\times 10^{-6}\ C$

Electrostatic force between charges, F = 9 N

Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$d=\sqrt{\dfrac{kq_1q_2}{F}}$

$d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}$

d = 0.71 meters

So, the distance between the charges is 0.71 meters. Hence, this is the required solution.

6 0

4 years ago

Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instant

Evgesh-ka [11]

Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5 ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

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3 years ago

What is the most destructive storm on earth

SIZIF [17.4K]

Hurricanes i believe

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3 years ago

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An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial:

$0.8\cdot t^{2}-56.92\cdot t +3.872 = 0$ (3)

$t_{1} \approx 71.082\,s$ , $t_{2}\approx 0.068$

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

3 0

3 years ago