Answer:
Tension T in each rope will be 254.56 N.
Explanation:
From the picture attached,
Weight suspended by the two ropes will be supported by the vertical components of the tension in the ropes.
Vertical components of both the ropes = Tsin(45)° + Tsin(45)°
= 2Tsin(45)°
=
=
Since,
T =
T = 254.56 N
Therefore, tension T in each rope will be 254.56 N.
Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:
Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,
Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m