A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ
1 answer:
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight
We replace in (1):
We solve the equation for m:
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data which is expressed in form of following way
here in above expression
= true value
= uncertainty in the value
now the relative uncertainty is given as
now by above formula we can say
a) 2.70 ± 0.05cm
here
True value = 2.70
uncertainty = 0.05
Relative uncertainty = = 0.0185
b) 12.02 ± 0.08cm
here
True value = 12.02
uncertainty = 0.08
Relative uncertainty = = 0.00665
Answer:
True
Explanation:
In this particular case, the area of the graph represents the impulse.
In fact, impulse is defined as the change in momentum of an object:
Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:
If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product
which corresponds to the definition of impulse.