A student leaves a sidewalk corner walks 3.0 blocks north then walks 3.0 blocks west.what is the students displacement from the
1 answer:
Answer:
The students displacement is 3·√3 blocks, 45° North of West
Explanation:
From the question, we have;
The distance North in which the student walks = 3.0 blocks
The next walking distance West by the student = 3.0 blocks
The displacement of the student = The shortest distance between the student start and final point
By representing the motion of the student in vector format, we have;
Therefore the magnitude of d is given as follows;
The direction of the student = tan⁻¹(3/3) = 45° West of North
The students displacement = 3·√3 blocks, 45° North of West.
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Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as
= 1.87 A