Explanation:
Neutralization reaction is a type of chemical reaction in which an acid and a base react together to form a salt <u><em>and water</em> </u>as products.
Explanation:
The dimensions of a standard backpack is 30cm x 30cm x 40cm
The mass of an average student is 70 kg
We know that, the density of gold is 19.3 g/cm³.
Let m be the mass of the backpack. So,

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.
Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
![F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7Bm_1m_2%7D%7BR%5E2%7D%3C%2Fp%3E%3Cp%3EThe%20problem%20states%20%20that%20%20the%20distance%20decrease%20to%201%2F10%20of%20the%20original%20distance%2C%20this%20means%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DR_2%20%3D%20%5Cfrac%7B1%7D%7B10%7DR)
And the force at this distance would be written in terms of the same equation:

Find the ratio between the final and the initial force:

Substitute the value for the final distance in terms of the initial distance:

Simplify:

This means the new force will be \frac{1}{100} of the original force.