Answer:
73.13°
Explanation:
According to snell's law,
n1sinθi = n2sinθr
n1/n2 = sinθr/sinθi
Critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°
This means i=C and r = 90°
The Snell's law formula will become
n1/n2 = sinC/sin90°
n2/n1 = 1/sinC
Where n1 is the refractive index of the less dense medium = 1.473
n2 is the refractive index of the denser medium = 1.540
Substituting the values in the formula,
1.540/1.473 = 1/sinC
1.045 = 1/sinC
SinC = 1/1.045
SinC = 0.957
C = sin^-1(0.957)
C = 73.13°
1. The problem statement, all variables and given/known data A person jumps from the roof of a house 3.4 meters high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 meters. If the mass of his torso (excluding legs) is 41 kg. A. Find his velocity just before his feet strike the ground. B. Find the average force exerted on his torso by his legs during deceleration. 2. Relevant equations I can't even seem to figure that part out. Help please? 3. The attempt at a solution I don't know how to start this at all
Answer:
When extra energy is added
Explanation:
When the ball is released from rest and swings back towards your face, it will only pass closer to the end of the nose as per the initial conditions. However, when extra energy is added to the ball, it strikes the nose since its velocity and heights are increased. Therefore, the only condition under which the ball hits your nose is when extra energy is added to the system.
Answer:
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from
The answer to the question is
The speed
of the electron when it is 10.0 cm from charge Q₁
= 7.53×10⁶ m/s
Explanation:
To solve the question we have
Q₁ = 3.45 nC = 3.45 × 10⁻⁹C
Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C
2·d = 50.0 cm
a = 10.0 cm
q = -1.6×10⁻¹⁹C
Also initial kinetic energy = 0 and
Initial electric potential energy = 
Final kinetic energy due to motion = 0.5·m·v²
Final electric potential energy = 
From the energy conservation principle we have

Solving for v gives

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg
gives v =7528188.32769 m/s or 7.53×10⁶ m/s
= 7.53×10⁶ m/s
Answer:
Note: Angular momentum is always conserved in a collision.
The initial angular momentum of the system is
L = ( It ) ( ωi )
where It = moment of inertia of the rotating circular disc,
ωi = angular velocity of the rotating circular disc
The final angular momentum is
L = ( It + Ir ) ( ωf )
where ωf is the final angular velocity of the system.
Since the two angular momenta are equal, we see that
( It ) ( ωi ) = ( It + Ir ) ( ωf )
so making ωf the subject of the formula
ωf = [ ( It ) / ( It + Ir ) ] ωi
Explanation: