Using F = Ke
F = 39200×(25÷100) = 39200×0.25
F= 9800N
)
5
-5
1 2 3
4
5
Other than at t = 0, when is the velocity of
the object equal to zero?
1. 5.0 s
2. 4.0 s
3. 3.5 s
4. At no other time on this graph. correct
5. During the interval from 1.0 s to 3.0 s.
Explanation:
Since vt =
Z t
0
a dt, vt
is the area between
the acceleration curve and the t axis during
the time period from 0 to t. If the area is above
the horizontal axis, it is positive; otherwise, it
is negative. In order for the velocity to be zero
at any given time t, there would have to be
equal amounts of positive and negative area
between 0 and t. According to the graph, this
condition is never satisfied.
005 (part 1 of 1) 0 points
Identify all of those graphs that represent motion
at constant speed (note the axes carefully).
a) t
x
b) t
v
c) t
a
d) t
v
e) t
a
Sound energy is produced when an object vibrates so an example would be a telephone ringing or someone playing a bass guitar
C because the air will push him closer to the space station
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