The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Answer:
2.77 * 10^5 m/s
Explanation:
Let us recall that kinetic energy is given by 1/2 mv^2
Where;
m = mass of the body
v = velocity of the body
In this case,
m = 3.38 * 10^31 kg
KE= 1.30 * 10^42 J
KE = 1/2 mv^2
v = √2KE/m
v = √2 * 1.30 * 10^42/3.38 * 10^31
v = √7.69 * 10^10
v = 2.77 * 10^5 m/s
Answer:
x = 1474.9 [m]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.
We must understand that when forces are applied on the body, they tend to slow the body down to stop it.
So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.
ΣF = m*a
![10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]](https://tex.z-dn.net/?f=10%20%2B12%2Asin%2860%29%3D%20-%206%2Aa%5C%5Ca%20%3D%20-%203.39%5Bm%2Fs%5E%7B2%7D%5D)
Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

where:
Vf = final velocity = 0 (the block stops)
Vo = initial velocity = 100 [m/s]
a = - 3.39 [m/s²]
x = displacement [m]
![0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]](https://tex.z-dn.net/?f=0%20%3D%20100%5E%7B2%7D-2%2A3.39%2Ax%5C%5Cx%3D%5Cfrac%7B10000%7D%7B2%2A3.39%7D%5C%5Cx%3D1474.9%5Bm%5D)
Answer:
1. As rocket mass increases, acceleration decreases.
2. The inverse of the mass of the boat.
Explanation:
1. Newton's second law of motion states;
F = ma
where F is the force applied, m is the mass and a is the acceleration.
Therefore, increasing the mass of a rocket increases its weight which would reduce its acceleration provided that the force is constant. Thus, as rocket mass increases, acceleration decreases.
2. The slope of the graph can be expressed as;
From Newton's second law,
F = ma
Slope = (Δa) ÷ (ΔF)
Slope = 
⇒
= 
Therefore, the slope of the graph is the reciprocal of the mass of the boat.
Answer:
New volume, v2 = 0.8L
Explanation:
<u>Given the following data;</u>
Original Volume = 2L
Original Temperature = 280K
New Temperature = 700K
To find new volume V2, we would use Charles' law.
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
Making V2 as the subject formula, we have;


V2 = 0.8L
Therefore, the volume of the gas after it is heated is 0.8L.