Kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
2 answers:
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There are already 2 givens, hence we can do direct substitution to get the answer. To make the process simpler, derive the distance formula from the Work formula.
Work = Force x Distance
Distance =
Work is 450J while the force is 150N hot tub
To get the proper units, get the equivalent of Joule to eliminate newton. A joule Is equal to 1 N-m
Distance =
Distance = 3m
Hence, the hot tub is lifted 3 meters.
Work = Force x Distance
Distance =
Work is 450J while the force is 150N hot tub
To get the proper units, get the equivalent of Joule to eliminate newton. A joule Is equal to 1 N-m
Distance =
Distance = 3m
Hence, the hot tub is lifted 3 meters.
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