An astronomer observes an asteroid in the solar system. He notes that the asteroid is three times farther from the Sun than Eart
2 answers:
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Answer:
The acceleration of the electron is 1.457 x 10¹⁵ m/s².
Explanation:
Given;
initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s
distance traveled by the electron, d = 0.01 m
final velocity of the electron, v = 5.4 x 10⁶ m/s
The acceleration of the electron is calculated as;
v² = u² + 2ad
(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a
(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²
(2 x 0.01)a = 2.91375 x 10¹³
Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².
a)
We use the formula :
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the values in:
4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f
Calculating this we get:
32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f
Rearrange for v2f:
v2f =
This gives us 8.0 m/s as the final velocity of the second ball.
b)
Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.
This means we use the formula:
Ek = +
=
+
Substituting in values:
Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2
This simplifies to:
Ek= 128J + 0J = 0J + 128J
This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.