Answer:
The heavier piece acquired 2800 J kinetic energy
Explanation:
From the principle of conservation of linear momentum:
0 = M₁v₁ - M₂v₂
M₁v₁ = M₂v₂
let the second piece be the heavier mass, then
M₁v₁ = (2M₁)v₂
v₁ = 2v₂ and v₂ = ¹/₂ v₁
From the principle of conservation of kinetic energy:
¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J
¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400
¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400
K.E₁ + ¹/₂K.E₁ = 8400
Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁
1.5 K.E₁ = 8400
K.E₁ = 8400/1.5
K.E₁ = 5600 J
K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J
Therefore, the heavier piece acquired 2800 J kinetic energy
Answer:
The value of each charge is 4.22 x 10⁻⁵ C
Explanation:
Given;
distance between the two identical charges, d = 2 m
the force of repulsion between these two charges, F = 4N
Apply Coulomb's law;
![F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D%20%5C%5C%5C%5Cbut%20%5C%20q_1%20%3Dq_2%2Cthen%20%5C%20let%20%5C%20q_1%20%3Dq_2%20%3D%20q%5C%5C%5C%5CF%20%3D%20%5Cfrac%7Bkq%5E2%7D%7Br%5E2%7D%5C%5C%5C%5Cq%5E2%20%3D%20%5Cfrac%7BFr%5E2%7D%7Bk%7D%5C%5C%5C%5Cq%5E2%20%3D%20%5Cfrac%7B4%2A2%5E2%7D%7B9%2A10%5E9%7D%20%5C%5C%5C%5Cq%20%5E2%20%3D%201.7778%2A10%5E%7B-9%7D%5C%5C%5C%5Cq%20%3D%20%5Csqrt%7B1.7778%2A10%5E%7B-9%7D%7D%5C%5C%5C%5Cq%20%3D4.22%20%2A10%5E%7B-5%7D%20C%5C%5C%5C%5Cq%3D%20q_1%3Dq_2%3D%204.22%20%2A10%5E%7B-5%7D%20C)
Therefore, the value of each charge is 4.22 x 10⁻⁵ C
Answer:
The acceleration is 6 [m/s^2]
Explanation:
We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.
![v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2B%20a%2At%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%20%3D%20final%20velocity%20%3D%2022%20%5Bm%2Fs%5D%5C%5Cv_%7Bi%7D%20%3D%20initial%20velocity%20%3D%204%20%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%3D%203%20%5Bs%5D%5C%5C)
Now replacing the values we have:
![a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_%7Bf%7D%20-%20v_%7Bi%7D%20%7D%7Bt%7D%20%5C%5Ca%3D%5Cfrac%7B22%20-%204%20%7D%7B3%7D%5C%5Ca%20%3D%206%20%5Bm%2Fs%5E%7B2%7D%20%5D)
Elements with atomic numbers from 58 through 71 are part of the
<span>
lanthanide</span> series <span />