When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.
I hope this helps. :)
Brainliest?
I think you can just sub the values in? unless the qn is asking for smth else?
M1U1 + M2V2 = (M1+M2)V, where M1 is the mass of the moving car, M2 is the mass of the stationary car, U1 is the initial velocity, and V is the common velocity after collision.
therefore;
(1060× 16) + (1830 ×0) = (1060 +1830) V
16960 = 2890 V
V = 5.869 m/s
The velocity of the cars after collision will be 5.689 m/s
Answer:
Explanation:
kinetic energy required = 1.80 MeV
= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 2.88 x 10⁻¹³ J
If v be the velocity of proton
1/2 x mass of proton x v² = 2.88 x 10⁻¹³
= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³
v² = 3.45 x 10¹⁴
v = 1.86 x 10⁷ m /s
If V be the potential difference required
V x e = kinetic energy . where e is charge on proton .
V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³
V = 1.8 x 10⁶ volt .
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
