Answer:
The moment of inertia is 0.7500 kg-m².
Explanation:
Given that,
Mass = 2.2 kg
Distance = 0.49 m
If the length is 1.1 m
We need to calculate the moment of inertia
Using formula of moment of inertia

Where, m = mass of rod
l = length of rod
x = distance from its center
Put the value into the formula


Hence, The moment of inertia is 0.7500 kg-m².
Answer:
A) θ = 13.1º , B) E
Explanation:
A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.
In this reference system the only force that we must decompose is the Normal one, let's use trigonometry
sin θ = Nₓ / N
cos θ =
/ N
Nₓ = N sin θ
Ny = N cos θ
x-axis (radial)
Nₓ = m a
where the acceleration is centripetal
a = v² / R
we substitute
-N sin θ = -m v² / R (1)
the negative sign indicates that the force and acceleration towards the center of the circle
y-axis (Vertical)
Ny - W = 0
N cos θ = mg
N = mg / cos θ
we substitute in 1
mg / cos θ sin θ = m v² / R
g tan θ = v² / R
θ = tan⁻¹ (v² / gR)
we calculate
θ = tan⁻¹ (25² / 9.8 274)
θ = 13.1º
B) when comparing the equations the correct one is E
Answer:
v = √2G
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1
/ R = - G m1
/ R
v² = 2G
(1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G
/ R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1
/ R
Em = - G m1
/ R
R = int ⇒ Em = 0
The answer is c Ai3n3 yea that’s the right answer idk
Answer:
80 J
Explanation:
Ep = mgh
Ep = (4 kg) (10 m/s²) (2 m)
Ep = 80 J