Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
A. an accelerating charged charged particle or changing magnetic fields
Answer:
Explanation:
Here's what we know because it was given to us:
a = -9.8 m/s/s and
time = 3.32 seconds
Here's what we know because we rock physics:
v₀ = 0 (because the object was held still before it was dropped).
Here's the equation that ties all that info together in a single one-dimensional equation:
v = v₀ + at
Filling in and solving for v:
v = 0 + (-9.8)(3.32) and
v = -33m/s
The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)
