Assume a 2D plane and both the birds start at origin. +X is east, -X is west, +Y is North, -Y is South.
So, after travel position of bird1 is (-2.8, 2.7) and bird2 is (-2.7,2.8);
slope of bird1 is 2.8/2.7
slope of bird2 is 2.7/2/8
Now we have two solpes b1 and b2
formula = (b2-b1)/(1+b1*b2);
Answer:
Coil 2 have 235 loops
Explanation:
Given
The number of loops in coil 1 is n
₁=
159
The emf induced in coil 1 is ε
₁
=
2.78
V
The emf induced in coil 2 is ε
₂
=
4.11
V
Let
n
₂ is the number of loops in coil 2.
Given, the emf in a single loop in two coils are same. That is,
ϕ
₁/n
₁=
ϕ
₂
n
₂⟹
2.78/159
=
4.11/
n
₂
n₂=
n₂=235
Therefore, the coil 2 has n
₂=
235 loops.
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
