Maybe the water wasnt stable enough and probably couldnt read the water level correctly
Answer:
Explanation:
Given that,
The compression in the spring, x = 0.0647 m
Speed of the object, v = 2.08 m/s
To find,
Angular frequency of the object.
Solution,
We know that the elation between the amplitude and the angular frequency in SHM is given by :
A is the amplitude
In case of spring the compression in the spring is equal to its amplitude
So, the angular frequency of the spring is 32.14 rad/s.
Answer:
182 to 3 s.f
Explanation:
Workdone for an adiabatic process is given as
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
where γ = ratio of specific heats. For carbon dioxide, γ = 1.28
For an adiabatic process
P₁V₁ʸ = P₂V₂ʸ = K
K = P₁V₁ʸ
We need to calculate the P₁ using ideal gas equation
P₁V₁ = mRT₁
P₁ = (mRT₁/V₁)
m = 2.80 g = 0.0028 kg
R = 188.92 J/kg.K
T₁ = 27°C = 300 K
V₁ = 500 cm³ = 0.0005 m³
P₁ = (0.0028)(188.92)(300)/0.0005
P₁ = 317385.6 Pa
K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
V₁ = 0.0005 m³
V₂ = 2.10 dm³ = 0.002 m³
1 - γ = 1 - 1.28 = - 0.28
W =
18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)
W = -67.47 (5.698 - 8.4)
W = 182.3 = 182 to 3 s.f
Answer:
a) p = 4.96 10⁻¹⁹ kg m / s
, b) p = 35 .18 10⁻¹⁹ kg m / s
,
c) p_correst / p_approximate = 7.09
Explanation:
a) The moment is defined in classical mechanics as
p = m v
Let's calculate its value
p = 1.67 10⁻²⁷ 0.99 3. 10⁸
p = 4.96 10⁻¹⁹ kg m / s
b) in special relativity the moment is defined as
p = m v / √(1 –v² / c²)
Let's calculate
p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)
p = 4.96 10⁻¹⁹ / 0.141
p = 35 .18 10⁻¹⁹ kg m / s
c) the relationship between the two values is
p_correst / p_approximate = 35.18 / 4.96
p_correst / p_approximate = 7.09
