Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
Answer:
0.631 m
6.53315 J
Explanation:
m = Mass = 2.47 kg
v = Velocity = 2.30 m/s
k = Spring constant = 32.8 N/m
A = Amplitude
In this system the energy is conserved
The amplitude is 0.631 m
Mechanical energy is given by
The mechanical energy is 6.53315 J
Answer:
the angular acceleration is 9.7 rad/
Explanation:
given information:
mass of thin rod, m = 3.2 kg
the length of the rod, L = 1.2
angle, θ = 38
to find the acceleration of the rod, we can use the torque's formula as below,
τ = Iα
where
τ = torque
I = inertia
σ = acceleration
moment inertia of this rod, I
I =
τ = F d, d = cosθ
τ = m g cosθ
now we can substitute the both equation,
τ = Iα
α = τ/I
= (m g cosθ)/(
)
= 3gcosθ/2L
= 3 (9.8)cos 38°/(2 x 1.2)
= 9.7 rad/
d.Both points have the same instantaneous angular velocity.
Explanation:
Let's analyze each option one by one:
a.the angular velocity at point A is greater than that of point B. --> FALSE. In fact, for a rigid body (such as the rigid wheel in this problem), all the points along the body rotates with the same angular velocity. This is because the angular velocity is defined as
where is the angular displacement and t is the time taken. Therefore, all the points rotate by the same angle
in the same time t.
b.Both points have the same centripetal acceleration. --> FALSE. The centripetal acceleration of a point along the disk is given by
where is the angular velocity and r is the distance of the point from the axis of rotation. Here A is located farther away from the axis with respect to B, so it will have a greater centripetal acceleration.
c.Both points have the same tangential acceleration. --> FALSE. The tangential acceleration is given by
where is the angular acceleration.
is the same for every point for a rigid body, but
is not (A is located farther away), so A and B do not have same tangential acceleration.
d.Both points have the same instantaneous angular velocity. --> TRUE. This is true for what we said at point A).
e.Each second, point A turns through a greater angle than point B. --> FALSE. The two points cover the same angular displacement in the same time, since they have same angular velocity.
Learn more about circular motion:
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