Answer:
If you know that that free fall acceleration g on the Moon is about 6 times less than on the Earth, it gives you the answer: on the Moon the same pendulum will have a period about √6≈2.45 longer than on the Earth.
Answer:
The tension of the string is 41.876 N
Explanation:
Given;
length of the string, L = 2.11 m
mass of the string, m = 19.5 g = 0.0195 kg
frequency of the wave, f = 440 Hz
wavelength, λ = 15.3 cm = 0.153 m
The velocity of the wave is given by;
v = fλ
v = 440 x 0.153
v = 67.32 m/s
Also the velocity of the wave is given by
where;
μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m
T is the tension of the string
T = v²μ
T = (67.32)²(0.00924)
T = 41.876 N
Therefore, the tension of the string is 41.876 N
Explanation:
Hey there!!
Let's simply work with it.
Here,
load = 1200N
Effort = 200N
Load distance = 15cm
We have,
According to the principle of lever.
L×LD = E×ED.
1200×15 = 200× ED.
18000 = 200ED.
Therefore, Effort Distance = 90cm.
<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
For transverse waves, the waves move in perpendicular direction to the source of vibration. For longitudinal waves, the waves move in parallel direction to the source of vibration . They are similar in the sense that energy is transferred in the form of waves.
