A drunken sailor stumbles 550 meters north, 500 meters northeast, then 450 meters northwest. What is the total displacement and
1 answer:
Answer:
Resultant displacement = 1222.3 m
Angle is 88.3 degree from +X axis.
Explanation:
A = 550 m north
B = 500 m north east
C = 450 m north west
Write in the vector form
A = 550 j
B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j
C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j
Net displacement is given by
R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j
R = 35.4 i + 1221.8 j
The magnitude is
The direction is given by
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Answer:
This is the answer of your question.
(a)
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
And the average force on the person is given by
with m = 75.0 kg being the mass of the person. Substituting,
where the negative sign means the force is opposite to the direction of motion of the person.
b)
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
So the average force on the person is
Answer:
a) I = 1,944 10⁻⁴⁶ Kg m²
, b) I = 7,915 10⁻⁵¹ Kg m²
Explanation:
a) The moment of inertia of point masses is
I = m r²
The nuclei have a very small size (10-15 m) so we can consider them punctual.
The distance to the center of mass passing through the middle of the two nuclei of equal mass
r = 0.6 10⁻¹⁰ m
The moment of total inertia
I = I₁ + I₂ = 2 I₀
I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²
I = 1,944 10⁻⁴⁶ Kg m²
The kinetic energy of the rotation is
w = h / 2π
K = ½ I w²
K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²
K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)
K = 2.16 10⁻⁹⁵ eV
B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem
I = + m R²
I = r² +
R²
Where R we calculate it by Pythagoras
R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2
R = √ (0.61 10⁻²⁰)
R = 0.78 10⁻¹⁰ m
I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²
I = (2,275 +5.54) 10⁻⁵¹
I = 7,915 10⁻⁵¹ Kg m²
The rotation energy of the electron
K = ½ I w²
If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.