Answer:
261.3 m/s
Explanation:
Mass of bullet=m=15 g=
1 kg=1000g
Mass of block=M=3 kg
d=0.086 m
Total mass =M+m=3+0.015=3.015 kg
K.E at the time strike=Gravitational potential energy at the end of swing

Using g=
Substitute the values




Velocity after collision=V=1.3 m/s
Velocity of block=v'=0
Using conservation law of momentum

Using the formula




Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.

R = 0.18 m
Answer:
the angle of incidence θ is 45.56 º
Explanation:
Given data
strikes the mirror before wall x = 30.7 cm
reflected ray strikes the wall y = 30.1 cm
to find out
the angle of incidence θ
solution
let us consider ray is strike at angle θ so after strike on surface ray strike to wall at angle 90 - θ
we will apply here right angle triangle rule that is
tan( 90 - θ) = y /x
tan( 90 - θ) = 30.1 / 30.7
90 - θ = tan^-1 (30.1/30.7)
90 - θ = 44.4345
θ = 45.56 º
the angle of incidence θ is 45.56 º