Consider a person sliding down a water slide at constant velocity. What are the forces acting on the person as they slide? Deter
1 answer:
Answer:
<em>D) Gravity and friction are acting on the person. The forces are balanced. There is no net force because the person is moving at a constant velocity</em>
Explanation:
<u>Net Force</u>
The first Newton's Law explains that a body keeps its velocity or at rest unless a net force is exerted on it. The second Newton's law gives us a mathematical expression for that net or unbalanced force as:
where m is the mass of the object and a its acceleration. Recall a body at constant velocity has zero acceleration.
The question describes how a person is sliding down a water slide at a constant velocity. That's all we need to know there is no net force acting on it. All the forces are balanced. The only possible options supporting this statement are B and D. We don't know if there is air resistance (B) or friction (D), but option B suggests the forces are balanced until the person reaches the bottom of the water slide. Option D is correct because the forces are balanced all the time
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Answer:
(a) 1.87 x 10⁻⁴ m/s
(b) 0.013V
Explanation:
(a) Drift speed, , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;
=
----------------(i)
Where;
q = amount of charge
n = free charge density
A = cross-sectional area of the wire
But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e
J = =
Equation (i) can then be written as follows;
=
=
=
---------------------(ii)
From the question;
E = 0.0520N/C
p = 1.72 x 10⁻⁸ Ωm
n = 8.5 x 10²⁸ electrons/m³
c = charge on electron = 1.9 x 10⁻¹⁹C
Substitute these values into equation (ii) as follows;
=
= 1.87 x 10⁻⁴ m/s
(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e
V = E x d [where d = 25.0cm = 0.25m]
V = 0.0520 x 0.25
V = 0.013V