22. reduction
25. Le Chatelier's principle
Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course :
= 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) ×
)
0 = 129.96 - 2.3
2.3
= 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √(
² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
hello your question is incomplete attached below is the complete question
answer :
a) I1 = I2
b) J1 > J2
c) E 1 > E2
d) ( vd1 ) > ( vd2 )
Explanation:
a) The currents in the two segments are the same i.e. I1 = I2 and this is because the segments are connected in series
b) Comparing the current densities J1 and J2 in the two segments
note : current density ∝ 1 / area
The area of the second segment is > the area of first segment therefore
J1 > J2
J1 ( current density of first segment )
J2 ( current density of second segment )
c) Comparing the electric field strengths E1 and E2
note : electric field strength ∝ current density
since current density of first segment is > current density of second segment and conductivity of the materials are the same hence
E 1 > E2
d) Comparing the drift speeds Vd1 and Vd2
( vd1 ) > ( vd2 )
this because ; vd ∝ current density