The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
<h3>
Time of motion of the girl</h3>
The time of motion of the girl is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- t is time of motion
- g is acceleration due to gravity
Substitute the given parameters and solve for time of motion;
50.8 = 0 + ¹/₂(9.8)t²
2(50.8) = 9.8t²
101.6 = 9.8t²
t² = 101.6/9.8
t² = 10.367
t = √10.367
t = 3.22 seconds
<h3>Final vertical velocity of the skydiver</h3>
vf = vi + gt
where;
vi is the initial vertical velocity = 0
vf = 0 + 9.8(3.22)
vf = 31.56 m/s
Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.
Learn more about vertical velocity here: brainly.com/question/24949996
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Answer:
D. has no overall force acting on it.
Explanation:
Why?
Because in a straight line at the constant speed means the car moving in the same velocity, which is not acceleration neither deceleration, and it cannot be on a downhill slope. So the correct answer is
<h3>→ D. has no overall force acting on it.</h3>
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
Iron...................... hope this helpes
