Answer:
B
Explanation:
the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating
Hope this helps :)
1. a. longitudinal waves.
There are two types of waves:
- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave
- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.
So, these types of waves are called longitudinal waves.
2. d. a medium
There are two types of waves:
- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)
- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium
3. a. AM/FM radio
Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.
Answer:
5 ms-2
Explanation:
F = ma
F = 100N
m = 20kg ( you should make sure the unit is kg before you answer the question)
100 = 20a
a = 100÷ 20
a = 5 ms-2
Answer:
0.006<357<700.003<6010<9256.0<9520.00
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
