i think the answer is B-House Rules Comittee
Answer:
The frog takes 8 jumps to reach top of well
Explanation:
Given data
Frog at bottom=17 foot
Each time frog leaps 3 feet
Frog has not reached the top of the well, then the frog slides back 1 foot
To Find
Total number of leaps the frog needed to escape from well
Solution
in 1 jump distance jumped=3+(-1)
=2 feet
=2×1 feet
The "-1" is because the frog goes back
Now After 2 jumps the distance jumped as:
Distance Jumped=2+2
Distance Jumped=2*2
=4 feet
Similarly after 7 jumps
Distance Jumped=2+2+......+2
Distance Jumped=2*7
=14 feet
Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.
So
Distance Jumped=(Distance Jumped after 7 jumps)+3
=14+3
=17 feet
The frog takes 8 jumps to reach top of well
Answer:
0.52 Nm
Explanation:
A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T
Angle between the plane of loop and magnetic field = 30 Degree
Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree
θ = 60°
Torque = N i A B Sinθ
Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60
Torque = 0.52 Nm
Answer:
C. over land in polar regions.
Answer:
i. The radius 'r' of the electron's path is 4.23 × 
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ = 
(since it enters perpendicularly to the field), q = e = 1.6 × 
C and m = 9.11 × 
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 × × 1.63 × )
= 1.10231 × 
÷ 2.608 × 
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 × × 1.63 × ) ÷ (2 ×
× 9.11 × )
= 2.608 × ÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
