What is the Nearest Galaxy from the Milky Way
2 answers:
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Answer:
Torque,
Explanation:
It is given that,
Force acting on the particle,
Position of the particle,
We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :
The cross product of vectors is given by :
or
So, the torque on the particle about the origin . Hence, this is the required solution.
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
