Answer:
The answer is 3,064x
Explanation:
When the collision happens, the momentum of the first car is applied to the both of them.
So we can calculate the force that acts on both cars as:
- The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/s
- The acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s
- Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.
Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:
- F = (2020+2940) x 5.78 = 28,684
The friction force acts in the opposite direction and if they stop after moving 2.12 meters;
- Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.
- F - Ffriction = m x V
- 28,684 - μ x 49,600 = 4960 x 5.78
- μ = 3,064x
Answer:
1/9 of that just outside the smaller sphere
Explanation:
The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:
where
k is the Coulomb's constant
Q is the charge on the sphere
r is the distance from the centre of the sphere
Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is
The second sphere has a radius which is 3 times that of the smaller sphere:
So, the electric field just outside the second sphere is
So, the correct answer is 1/9.
The answer is B because it's 80 percent copper, that's a greater number then 20.
<span>The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.
So there will be only 2 electrons in the outer most shell.
According to the information mentioned above we can conclude the </span><span>unknown element likely belongs to the second group.
</span><span>I2 = 1752 kj/mol</span>
