Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation

angular frequency ω = 2π / T
= 
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.
Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.
Answer:
the change in thermal energy of the projectile is 43.8 kJ
Explanation:
Given;
mass of the object, m = 5kg
initial velocity of the projectile, v₁ = 200 m/s
final velocity of the projectile, v₂ = 150 m/s
To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.
Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²
KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)
KE = ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ
Therefore, the change in thermal energy of the projectile is 43.8 kJ
The possible resulting chemical formulas for an ionic compound with calcium given the respective charges of the ions are: CaO, CaMg, or CaF₂ and CaO, CaF₂, or CaCl₂. This is because when dealing with these compounds, you simply need to interchange the oxidation state of the two elements and place as the subscript of the element. For instance, when we have Ca²⁺ and F⁻, the result is CaF₂. However, when the oxidation states of the two compounds are equal, the subscript is 1. That is, for Ca²⁺ and Mg²⁻, the result is CaMg. And for Ca²⁺ and Cl⁻, the result is CaCl₂.