Answer:
Explanation:
Since this is a distance v time graph, the slope of the line from 1s to 3s is the velocity. However, it looks like, at t=3, the velocity is 0, so getting the definite velocity is not going to happen. We can estimate it as closely as possible. Since the line is tending from the upper left to the lower right, the slope is negative, so the velocity is also negative. That leaves only C or D as our answers. And the slope is closer to -1 than to -5, so choice D. is the one you want.
Answer:
You are doing this wrong. Make sure that if you have 2 rows and on the top it has pounds then on the second bottom row you also have pounds so that they cancel each other out.
Explanation:
Answer : When we increase the temperature of an exothermic reaction the equilibrium will shift to the left direction i.e, towards the reactant.
Explanation :
Le-Chatelier's principle : This principle states that if any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
As the given reaction is an exothermic reaction in which the heat is released during a chemical reaction. That means the temperature is decreased on the reactant side.
For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.
If the temperature is increases in the equilibrium then the equilibrium will shift in the direction where, temperature is getting decreased. Thus, the reaction will shift to the left direction i.e, towards the reactant.
Hence, when we increase the temperature of an exothermic reaction the equilibrium will shift to the left direction i.e, towards the reactant.
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.