Answer:
A . Are the masses repelling or attracting
answer:
<em><u>So, positive energy densities ("masses") attract each other by gravitational interaction. This is the general idea.</u></em>
<em><u>B</u></em>. What is the magnitude of the electrical force between the objects<em><u>?</u></em>
<em><u>answer</u></em><em><u>:</u></em>
<em><u> </u></em><em><u>The magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.</u></em>
C. What would the magnitude of the force be if one of the charges was 1/3 the amount?
answer:
<em><u>0.45</u></em>
D. ) What would the force be if the distance between the charges was only 1 meter (with the original charges)?
answer;
<em><u>Fmin=2.3×10^−28</u></em><em><u>N</u></em>
<em><u>E</u></em><em><u>:</u></em><em><u> </u></em>What would the force be if one of the charges was 1/3 the amount AND the distance was 1 meter?
answer;
<em><u>Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.</u></em>
<em><u>#</u></em><em><u>CARRYONLEARNING</u></em><em><u>:</u></em><em><u>)</u></em>
<em><u>LOVEUALL</u></em>
Answer:
k_2 = 7.815 * 10^-3 s^-1
Explanation:
Given:
- rate constant of reaction k_1 = 7.8 * 10^-3 s^-1 @ T_1 = 25 C
- rate constant of reaction k_2 = ? @ T_2 = 75 C
- The activation energy E_a = 33.6 KJ/mol
- Gas constant R = 8.314472 KJ / mol . K
Find:
- rate of reaction k_2 @ T_2 = 75 C
Solution:
- we will use a combined form of Arrhenius equations that relates rate constants k as function of E_a and temperatures as follows:
k_2 = k_1 * e ^ [(E_a / R) * ( 1 / T_1 - 1 / T_2 )
- Evaluate k_2 = 7.8 * 10^-3* e^[(33.6 / 8.314472)*(1/298 -1/348)
- Hence, k_2 = 7.815 * 10^-3 s^-1
We are asked to compare two series circuits having equal number of light bulbs.
1st circuit is powered by 6 batteries each having a voltage of 1.5V
2nd circuit is powered by a single battery having a voltage of 9V.
The six batteries in the 1st circuit can be connected together in series or in parallel.
When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means
Voltage of pack = number of batteries*voltage of each battery
Voltage of pack = 6*1.5
Voltage of pack = 9 volts
But the current remains same in the series connection since there is only path for the current to flow.
On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.
Circuit 1:
In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Circuit 2:
In this circuit, we have 1 battery which provide a voltage of 9 volts.
We have connected 2 bulbs in this series circuit just like in circuit 1.
The voltage will be equally divided between two bulbs if both bulbs are identical in construction.
So there will be 4.5 volts across each bulb and both bulbs will have same brightness.
Conclusion:
Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.
Friction as it will move charge (electrons) from one object to another
More water evaporates than falls as rain
did u know 3 million tonnes of water gets evaporated from oceans per day