Momentum
mava + mbvb = mava '+ mbvb'
(300 x 10)+(150 x 0) = (300 x 4.12)+(150 x vb')
3000=1236+150vb'
1764 = 150vb'
vb'=+11.76 m/s ≈ +11.8 m/s (positive sign, to the right)
Answer:
The frequency of oscillation of the simple pendulum is 0.49 Hz.
Explanation:
Given that,
Mass of the simple pendulum, m = 0.35 kg
Length of the string to which it is attached, l = 1 m
We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.
Answer:
1.F is the electrostatic force between charges (in Newtons),
2.q₁ is the magnitude of the first charge (in Coulombs),
3.q₂ is the magnitude of the second charge (in Coulombs),
4.r is the shortest distance between the charges (in m),
5.ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C² .
Answer:
2.26 s
Explanation:
Let's take down to be positive.
Given (in the y direction):
Δy = 25 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
25 m = (0 m/s) t + ½ (9.8 m/s²) t²
25 = 4.9t²
t = 2.26 s
If the ball instead had an initial horizontal velocity of 5 m/s, its initial vertical velocity is still 0 m/s. So the time to fall is still 2.26 s.
Answer:
earth's shadow covering the moon,thats lunar eclipse