Something super duper uper stuper luper nuper tuper zuper yuper fuper guper huper kuper juper wuper special
Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
<span> B. A person moving a ball through a stream of water</span>
A) the periodic time is given by the equation;
T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
= 6.28 × √0.0673
= 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
thus; frequency = 0.614 Hz
b) The vertical distance, the height is given by
h= 0.66 cos 12
h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
mgh = 1/2mv²
v² = 2gΔh = 2×9.81 × 0.01
= 0.1962
v = 0.443 m/s
c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
= 1/2 × 0.310 × 0.443²
= 0.0304 J
Answer:
θ = Cos⁻¹[A.B/|A||B|]
A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result
Explanation:
We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:
A.B = |A||B| Cosθ
where,
A = 1st Non-Zero Vector
B = 2nd Non-Zero Vector
|A| = Magnitude of Vector A
|B| = Magnitude of Vector B
θ = Angle between vector A and B
Therefore,
Cos θ = A.B/|A||B|
<u>θ = Cos⁻¹[A.B/|A||B|]</u>
Hence, the correct answer will be:
<u>A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result</u>
