Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m
The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
- <em>Current flowing in the wire, I = 4.00 mA</em>
- <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
- <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
- <em>Length of wire, L = 2.00 m</em>
- <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>
<em />
The initial area of the copper wire;

The final area of the copper wire;

The initial drift velocity of the electrons is calculated as;

The final drift velocity of the electrons is calculated as;

The change in the mean drift velocity is calculated as;

The time of motion of electrons for the initial wire diameter is calculated as;

The time of motion of electrons for the final wire diameter is calculated as;

The average acceleration of the electrons is calculated as;

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
Learn more here: brainly.com/question/22406248
It is number 3 because I know it is
Answer:
V' = 0.84 m/s
Explanation:
given,
Linear speed of the ball, v = 2.85 m/s
rise of the ball, h = 0.53 m
Linear speed of the ball, v' = ?
rotation kinetic energy of the ball

I of the moment of inertia of the sphere

v = R ω
using conservation of energy


Applying conservation of energy
Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy



V'² = 0.7025
V' = 0.84 m/s
the linear speed of the ball at the top of ramp is equal to 0.84 m/s