Answer:
True
Explanation:
This is a representation of Gauss law.
Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.
Answer:
It is because it cannot be used time again and again.
Answer:
3.43 m/s^2
Explanation:
Force is equal to mass times acceleration. (F=ma). You can use inverse operations to get the formula for acceleration, which is acceleration is equal to force divided by mass. (a=F/m). Since there are two forces here, the force friction (55 N), and the force applied (175 N), we must solve for the net force. To solve for the net force, you take the applied force (175 N) and subtract the frictional force from it (55 N). Thus, the net force is 120 N. With this done, we can now solve for our acceleration.
Using the equation for acceleration, we take the force and divide it by mass.
120/35
Answer: 3.43* m/s^2**
*Note: This is rounded to the nearest hundredth, the full answer is: 3.42857143
**Note: In case you're confused, this is meters per second squared.
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :
Here, a = g
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
