Answer:
Check the explanation
Explanation:
(a) When the electric field is applied to the ends of the wire, electrons move through the wire because the resistance of the wire is less than the resistance of the surrounding air. Resistance is the obstruction to the flow of electrons hence electrons flow through the path having minimum resistance.
(b) As the thermal energy increases, Kinetic energy of electrons as well as that of positive ions also increase. Hence the mean time between collisions of electrons with ions is reduced due to movement of ions also. That is why Resistance of the wire increases.
The object that a satellite revolves around is the <em>central body</em> of the system. <em>(C)</em>
For example:
-- The central body of the solar system is the Sun.
-- The central body for TV satellites, GPS satellites, weather satellites, and the International Space Station is the Earth.
-- The central body for Phobos and Deimos is Mars.
This should be a pretty easy question to answer by elimination, when you notice that "Orbit", "Period", and "Rotation" are not "Bodies".
Answer:
<h2>4.9 J</h2>
Explanation:
The gravitational potential energy of a body can be found by using the formula
GPE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
GPE = 10 × 9.8 × 0.05
We have the final answer as
<h3>4.9 J</h3>
Hope this helps you
Answer:
Explanation:
We shall take the help of vector form of displacement . Taking east as i and north as j
4.0m N = 4 j
7.5 m E = 7.5 i
6.8 m S = - 6.8 j
3.7 m E, = 3.7 i
3.6 m S = - 3.6 j
5.3 m W = - 5.3 i
3.7 m N, = 3.7 j
5.6 m W = - 5.6 i
4.4 m S = - 4.4 j
4.9 m W = - 4.9 i
Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i
= -4.6 i -7.1 j
magnitude of displacement = 
= 8.46 m
Direction
Tanθ = 7.1/ 4.6
θ = 57⁰ south of west .
distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9
= 49.5 m
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
