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2 answers:
Answer:
1. Emma on top of the mountain
As she remains still and is elevated above ground, her energy is attributed to gravitational potential energy.
Thus, we have
gravitational potential energy
2. Emma leaping off the mountain
During free fall, Emma begins to descend with increasing speed, which means she has
motion energy
3. tension in the rope at Emma's lowest point
As the rope stretches and reaches its lowest point causing her speed to drop to zero, the energy is converted entirely to elastic potential energy.
elastic potential energy
4. Emma rebounding
When she bounces back, she regains kinetic energy, moving upwards with some speed.
motion energy
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Consider the diagram below.
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
The question pertains to the change in frequency of a wave noted by an observer moving in relation to the source, indicating that the concept to invoke is "Doppler's effect."
The standard formula for the Doppler effect is:
-- (A)
Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.
Where,
g = Speed of sound = 340m/s.
= Velocity of the observer relative to the medium =?.
= Velocity of the source in relation to the medium = 0 m/s.
= Frequency emitted from the source = 400 Hz.
= Frequency recognized by the observer = 408 Hz.
Substituting the given values into equation (A) will yield:
Solving the above will result in,
= 6.8 m/s
The correct result = 6.8m/s
The standard formula for the Doppler effect is:
Note that we don’t need to be concerned with the signs here, as all entities are moving toward each other. If something was moving away, a negative sign would apply, but that is not relevant to this scenario.
Where,
g = Speed of sound = 340m/s.
Substituting the given values into equation (A) will yield:
Solving the above will result in,
The correct result = 6.8m/s