Answer:
F = 0.535 N
Explanation:
Let's use the concepts of energy, at the highest and lowest point of the trajectory
Higher
Em₀ = U = mg y
Lower
= K = ½ m v²
Emo =
mg y = ½ m v2
v = √ 2gy
y = L - L cos θ
v = √ (2g L (1-cos θ))
Now let's use Newton's second law n at the lowest point where the acceleration is centripetal
F = ma
a = v² / r
In turning radius is the cable length r = L
F = m 2g (1-cos θ)
Let's calculate
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.
When polonium-210 emits an alpha particle, the child isotope has an atomic mass of </span><span> ⇒ 206</span>.</span>
<span><span>I-131 undergoes beta-minus decay. The chemical symbol for the new element is </span><span> ⇒ Xe</span>.</span>
<span><span>Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of </span><span> ⇒ 18</span>.</span>
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:
We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
![|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}](https://tex.z-dn.net/?f=%7Cv_%7Bhawk%7D%7C%3D%5Csqrt%5B%5D%7B6%5E2%2B2%5E2%7D%3D%5Csqrt%7B40%7D)
≈
And the angle with respect to the east should be with:
