Answer:
- <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>
Explanation
1. <u>Data</u>:
a) ω = constant
b) Abbie: r₁ = 1 m
c) Zak: r₂ = 2 m
d) Ac₁ = ? Ac₂
2. <u>Formulae</u>
3. <u>Solution</u>:
a) Abbie:
b) Zack:
c) Divide Ac₁ / Ac₂
- Ac₁ / Ac₂ = ω² (1m) / [ω² (2m) ] = 1/2
⇒ Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :



v = 2 m/s
Hence, their speed after collision is 2 m/s.
Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:

we've the density of the water 


b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:



<u>Now the difference in the column is :</u>


