Answer:
The magnitude of momentum of the airplane is
.
Explanation:
Given that,
Mass of the airplane, m = 3400 kg
Speed of the airplane, v = 450 miles per hour
Since, 1 mile per hour = 0.44704 m/s
v = 201.16 m/s
We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,
![p=mv](https://tex.z-dn.net/?f=p%3Dmv)
![p=3400\ kg\times 201.16 \ m/s](https://tex.z-dn.net/?f=p%3D3400%5C%20kg%5Ctimes%20201.16%20%5C%20m%2Fs)
![p=683944\ kg-m/s](https://tex.z-dn.net/?f=p%3D683944%5C%20kg-m%2Fs)
or
![p=6.83\times 10^5\ kg-m/s](https://tex.z-dn.net/?f=p%3D6.83%5Ctimes%2010%5E5%5C%20kg-m%2Fs)
So, the magnitude of momentum of the airplane is
. Hence, this is the required solution.
Answer:
True
Explanation:
This can be explained by the special theory of relativity for length contraction.
Length contraction is observed in the direction of motion of an object when an object moves with speed closer to the speed of light.
The length of the rocket in this case, appears shorter to the observer on earth in the stationary reference frame which is improper frame whereas the traveler in the rocket is in the same inertial frame which is proper for the rocket's size measurement.
Answer:
0.958 m
Explanation:
So the total mass of the system is
M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28 kg
let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation
![My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4](https://tex.z-dn.net/?f=My%20%3D%20m_1y_1%20%2B%20m_2y_2%20%2Bm_3y_3%20%2B%20m_4y_4)
![11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806](https://tex.z-dn.net/?f=11.28y%20%3D%201.93%2A2.91%20%2B%202.95%2A2.43%20%2B%202.41%2A0%20%2B%203.99%2A%28-0.496%29%20%3D%2010.806)
![y = \frac{10.806}{11.28} = 0.958 m](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B10.806%7D%7B11.28%7D%20%3D%200.958%20m)
So the center of mass is 0.958 m from the origin
Answer:
50 N
4.2 N
Explanation:
i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.
ii) When the boom is resting on the end support, the normal force is:
∑τ = Iα
-W (0.50) + F (3.0) − N (6.0) = 0
-0.50 W + 3.0 F = 6.0 N
N = (-0.50 W + 3.0 F) / 6.0
N = (-0.50 × 2350 + 3.0 × 400) / 6.0
N ≈ 4.2
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