Answer:
Verb. Load with too great a burden or cargo.
"Both boats were overloaded and low in the water"
Noun. An excessive load or amount.
"an overload of stress"
Explanation:
Similar words are strain, excess, and overburden.
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To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.
I will also attach a free body diagram that allows a better understanding of the problem.
For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore


Here,
m = Net mass
= Angular velocity
r = Radius
W = Weight
N = Normal Force

The net mass is equivalent to

Then,

Replacing we have then,

Solving to find the angular velocity we have,

Therefore the angular velocity is 0.309rad/s
Explanation:
The given data is as follows.
mass = 0.20 kg
displacement = 2.6 cm
Kinetic energy = 1.4 J
Spring potential energy = 2.2 J
Now, we will calculate the total energy present present as follows.
Total energy = Kinetic energy + spring potential energy
= 1.4 J + 2.2 J
= 3.6 Joules
As maximum kinetic energy of the object will be equal to the total energy.
So, K.E = Total energy
= 3.6 J
Also, we know that
K.E = 
or, v = 
= 
= 
= 6 m/s
thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.
Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
![t = 6.45 [light-years]*365[\frac{days}{1light-year}]*24[\frac{hr}{1day}] *60[\frac{min}{1hr}]*60[\frac{seg}{1min} ] =203407200 [s]](https://tex.z-dn.net/?f=t%20%3D%206.45%20%5Blight-years%5D%2A365%5B%5Cfrac%7Bdays%7D%7B1light-year%7D%5D%2A24%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%20%2A60%5B%5Cfrac%7Bmin%7D%7B1hr%7D%5D%2A60%5B%5Cfrac%7Bseg%7D%7B1min%7D%20%5D%20%3D203407200%20%5Bs%5D)
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]
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