In order to determine the angle of the refracted ray, we may apply Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media. Mathematically, this is:
n₁sin(∅₁) = n₂sin(∅₂)
Where n is the refractive index. Substituting the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91
The angle of the refracted ray is 15°.
Answer:
The magnitude of the net electric field is:
Explanation:
The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).
On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).
Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>
Let's find first E1 and E2.
The electric field equation is given by:
Where:
- k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
- q1 is the first charge
- d1 is the distance from q1 to P
And E2 will be:
Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.
I hope it helps you!
To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
The angular velocity of a body can be described as a function of frequency as
PART A) The expression for the maximum angular velocity is given by the amplitude so that
PART B) The maximum acceleration on your part would be given by the expression
you can subtract the atomic number from the mass number to find the number of neutrons.
Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²