While you are studying for an upcoming physics exam, a lightning storm is brewing outside your window. Suddenly, you see a tree
1 answer:
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Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Answer:
455165.278 m
Explanation:
P = Power = 3.7 W
v = Velocity = 10.7 m/s
Amount of fat = 4 g
1 gram of fat provides about 9.40 (food) Calories
Energy given by 4 g of fat
Time required to burn the fat
Distance traveled by the bird
The bird will fly 455165.278 m
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
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