Answer:

Explanation:
Given that,
The angular velocity of a wave, 
The maximum displacement of the wave, A = 10 cm (let)
The maximum acceleration of the wave is given by :

Put all the values,

So, the maximum acceleration of the wave is equal to
.
To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have





Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s
Answer:
The distance moved is 82 m.
The displacement is 18 m to the south.
Explanation:
The distance is a measure of the total length traveled along the path, while the displacement only takes into account the length between the starting position (departure) and final position (arrival). That is, distance refers to how much space an object travels during its movement, being the amount moved, while displacement refers to the distance and direction of the final position with respect to the initial position of an object.
So, the distance being the sum of the distances traveled, you get:
18 m + 22 m + 14 m + 28 m= 82 m
<u><em>The distance moved is 82 m.</em></u>
You know that the tennis ball moves 18 meters to the north, then 22 meters to the south, then 14 meters to the north, and finally 28 meters to the south. Then the tennis ball moves:
- northward: 18 m + 14 m= 32 m
- to the south: 22 m + 28 m= 50 m
Calculating the displacement as the difference between the final position and the initial position, you get:
displacement= 50 m - 32 m= 18 m
<u><em>
The displacement is 18 m to the south.</em></u>
Answer:
(D) V + 1.5 volts
Explanation:
Eq. for stopping potential:
eV₀ = hf - ∅ ---- eq (1)
where
V₀ = stopping potential
h = planks's constant
v = frequency of light
f = c/λ
∅ = work function
stopping potential for metal 1 =V₁= V
stopping potential for metal 1 = V₂
work function for metal 1 = 3.6 eVolts
work function for metal 2 = 2.1 eVolts
substituting values for metal 1 in eq (1)
eV₁ = hf - ∅₁
eV = hc/λ - 3.6 eVolts
V = hc/eλ - 3.6 Volts
hc/eλ = V + 3.6 Volts ----- eq(2)
Metal 2 is illuminated with same wavelength so stopping potential V₂ is:
eV₂ = hf - ∅₂
eV₂ = hc/λ - 2.1 eVolts
V₂ = hc/eλ - 2.1 Volts ---- eq(3)
substituting eq(2) in eq(3)
V₂ = V + 3.6 Volts - 2.1 Volts
V₂ = V + 1.5 Volts