Radio waves are the longest
Answer:
![F = 26.04 N](https://tex.z-dn.net/?f=F%20%3D%2026.04%20N)
Explanation:
As we know that system of two boxes are moving on frictionless surface
So here if two boxes are considered as a system
then we have
![F = (m_1 + m_2) a](https://tex.z-dn.net/?f=F%20%3D%20%28m_1%20%2B%20m_2%29%20a)
![m_1 = 24 kg](https://tex.z-dn.net/?f=m_1%20%3D%2024%20kg)
![m_2 = 62 kg](https://tex.z-dn.net/?f=m_2%20%3D%2062%20kg)
![F = 36 N](https://tex.z-dn.net/?f=F%20%3D%2036%20N)
![36 = (24 + 62) a](https://tex.z-dn.net/?f=36%20%3D%20%2824%20%2B%2062%29%20a)
![a = 0.42 m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.42%20m%2Fs%5E2)
Now since we know that both the boxes are moving together so force applied by first box on other box is given as
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
![F = (62 \times 0.42)](https://tex.z-dn.net/?f=F%20%3D%20%2862%20%5Ctimes%200.42%29)
![F = 26.04 N](https://tex.z-dn.net/?f=F%20%3D%2026.04%20N)
Answer:
Answer
Explanation:
A solid because each solids has various shapes and volume. scientist uses variety of tools such as, Cylinder,scaler, etc to differentiate the solids.
Answer:
The correct answer is A
Explanation:
The question requires as well the attached image, so please see that below.
Coulomb's Law.
The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges
and
, are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:
![|F|=k \cfrac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%20q_2%7D%7Br%5E2%7D)
Thus if we decrease the distance by half we have
![r_1 =\cfrac r2](https://tex.z-dn.net/?f=r_1%20%3D%5Ccfrac%20r2)
So we get
![|F|=k \cfrac{q_1 q_2}{r_1^2}](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%20q_2%7D%7Br_1%5E2%7D)
Replacing we get
![|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%20q_2%7D%7B%28r%2F2%29%5E2%7D%5C%5C%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%20q_2%7D%7Br%5E2%2F4%7D)
We can then multiply both numerator and denominator by 4 to get
![|F|=k \cfrac{4q_1 q_2}{r^2}](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7B4q_1%20q_2%7D%7Br%5E2%7D)
So we have
![|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)](https://tex.z-dn.net/?f=%7CF%7C%3D4%20%5Cleft%28k%20%5Ccfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%5Cright%29)
Thus if we decrease the distance by half we get four times the force.
Then we can replace the second condition
![q_{2new} =2q_2](https://tex.z-dn.net/?f=q_%7B2new%7D%20%3D2q_2)
So we get
![|F|=k \cfrac{q_1 q_{2new}}{r_1^2}](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%20q_%7B2new%7D%7D%7Br_1%5E2%7D)
which give us
![|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)](https://tex.z-dn.net/?f=%7CF%7C%3Dk%20%5Ccfrac%7Bq_1%202q_2%7D%7Br_1%5E2%7D%5C%5C%7CF%7C%3D2%5Cleft%28k%20%5Ccfrac%7Bq_1%20q_2%7D%7Br_1%5E2%7D%5Cright%29)
Thus doubling one of the charges doubles the force.
So the answer is A.