A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at
1 answer:
The velocity of the rocket is 7.8 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the rocket+fuel system must be conserved.
Before the launch, the total momentum of the system is zero, since the rocket and the fuel are at rest:
After the launch, the total momentum is
where
M = 4.0 kg is the mass of the rocket
V is the velocity of the rocket
m = 50.0 g = 0.050 kg is the mass of the fuel ejected
v = -625 m/s is the velocity of the fuel (taking "backward" as negative direction)
Since the total momentum is conserved, we have
So we can solve the equation to find V, the velocity of the rocket:
And the positive sign means the rocket moves forward.
Learn more about momentum:
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Answer:
(a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Explanation:
Given that,
Radius = 5.9 m
(a). Angle
We need to calculate the angle in radian
We need to calculate the path length
Using formula of path length
(b). Angle = 30 rad
We need to calculate the path length
(c). Angle = 30 rev
We need to calculate the angle in rad
We need to calculate the path length
Hence, (a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Answer:
elastic force and weight are related to the acceleration of the System.
Explanation:
The relationship between these two forces can be found with Newton's second law.
- W = m a
K x - m g = m a
We see that elastic force and weight are related to the acceleration of the System.
If a harmonic movement is desired, an extra force that increases the elastic force is applied, but to begin the movement this force is eliminated, in general , if the relationship between this external and elastic force is desired, the only requirement is that it be small for harmonic movement to occur