Answer:
vo = 227 m/s
Explanation:
Let's analyze the problem, we have two parts one during the crash for which we will use the moment and another when it slides where we will use the energy. As we give the data of the latter let's start here
Before starting let's reduce the unit to the SI system
m = 6.00 g (1kg / 1000g) = 6.00 10⁻³ kg
M = 1.12 kg
We write the energy at two points, one initial right after the crash and another when the body has stopped
Just after shock
The bodies are united, so the mass is the sum of the mass of the bullet and the block
Em₀ = K = ½ (m + M) v²
When the body has stopped
= 0
When in the system there is friction force the variation the mechanical energy is equal to the work of the friction force. Notice that the force of friction opposes the movement so that their work is negative
W fr = ΔEm = -Em₀
-fr d = 0 - ½ (m + M) v²
To find the force of friction let's use Newton's second law
Axis y
N-W = 0
N = W = (m + M) g
The equation for the force of friction is
fr = μ N
fr = μ (m + M) g
Let's replace in the work and energy equation
-μ (m + M) g d = 0 - ½ (m + M) v²
From here we can find the system speed. Let's calculate
v³ = 2 μ g d
v = √ (2μ gd)
v = √ (2 0.250 9.8 0.300)
v = 1.21 m / s
Now let's solve the crash, let's look for the moment before and after it
Before the crash
po = m vo
After the crash
= (m + M) v
The system is formed by the bullet and the block, therefore, the forces during the impact are internal and the amount of movement is conserved
po =
m vo = (m + M) v
vo = v (m + M) / m
vo = 1.21 (0.00600 + 1.12) /0.00600
vo = 227 m / s
The correct choice is
D. 22 Hz and 42 Hz.
In fact, the beat frequency is given by the difference between the frequencies of the two waves:
In this problem, the beat frequency is , therefore the only pair of frequencies that gives a difference equal to 20 Hz is
D. 22 Hz and 42 Hz.
Answer:
Explanation:
In this case, since the charged particle moves in circular motion, the centripetal force is equivalent to the magnetic force.
(a) 0.0021 s, 2926.5 rad/s
The frequency of the B note is
The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:
The angular frequency instead is given by
And substituting
f = 466 Hz
We find
(b) 20 Hz, 125.6 rad/s
In this case, the period of the sound wave is
T = 50.0 ms = 0.050 s
So the frequency is equal to the reciprocal of the period:
While the angular frequency is given by:
(c)
The minimum angular frequency of the light wave is
so the corresponding frequency is
and the period is the reciprocal of the frequency:
The maximum angular frequency of the light wave is
so the corresponding frequency is
and the period is the reciprocal of the frequency:
(d)
In this case, the frequency is
So the period in this case is
While the angular frequency is given by