Solve for the linear/tangential speed:
<em>a</em> = <em>v</em>²/<em>r</em>
where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.
4.7 m/s² = <em>v</em>²/(0.3 m)
<em>v</em>² = (0.3 m) (4.7 m/s²)
<em>v</em> ≈ 3.96 m/s
For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is
(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s
Take the reciprocal of this to get the period:
1 / (7.46 rev/s) ≈ 0.134 s/rev
So it takes the record about 0.134 seconds to complete one revolution.
Answer:
(a) Position Vectors V₁= -2î km, V₂=5î km
(b) Displacement Δx=7 km
Explanation:
Given data
Distance=2 km west at t=0
Distance=5 km east at t=6 min
Positive x is the east direction
To find
(a)Car position vector at given times
(b)Displacement between 0 to 6.0 min
Solution
For Part (a) car position vector at given times
At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁
V₁= -2î km
At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂
V₂=5î km
For (b) displacement between 0 to 6.0 min
According to following mathematical law we can conclude that
Δx=x₂-x₁
Δx=5-(-2)km
Δx=7 km
Answer:
The phase difference between these two waves is 141.1⁰
Explanation:
The displacement of the wave is given as;
Amplitude, A = 2yₓCos(¹/₂Φ)
Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A
A = 2(1.5A)Cos(¹/₂Φ)
A = 3ACos(¹/₂Φ)
¹/₃ = Cos(¹/₂Φ)
(¹/₂Φ) = Cos ⁻(0.3333)
(¹/₂Φ) = 70.55°
Φ = 141.1°
The phase difference between these two waves is 141.1⁰
Answer:
The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
Explanation:
Given that,
Mass of the meter stick, m = 0.3 kg
Center of mass is located at its 45 cm mark.
We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :
or
So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
