Answer:
2.24 m/s²
Explanation:
Using equation of motion
s = ut + at²
u = 0 , t = 3.17 s , s = 11.26 m
Put these values in the equation above
11.26 = 0 +.5 x a( 3.17)²
a = 2.24 ms⁻².
So acceleration due to gravity on that planet will be 2.24 m s⁻².
Answer:
(a) 1767.43 N
(b) 182.45 N
Explanation:
Radius of earth, R = 6450 km
Weight of person, W = 7070 N
mass of person, m = W / g = 7070 / 9.8 = 721.4 kg
(a) h = 6450 km
The value of acceleration due to gravity on height is given by
g' = g / 4 = 9.8 / 4 = 2.45 m/s^2
The weight of the person at such height is
W' = m x g' = 721.4 x 2.45
W' = 1767.43 N
(b) h = 33700 km
The value of acceleration due to gravity on height is given by
g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2
The weight of the person at such height is
W' = m x g'
W' = 721.4 x 0.253
W' = 182.45 N
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