The number of protons determines the element. Many elements also have many isotopes. Such as carbon-14 Carbon-12 and carbon-10.
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The force result in stretching the spring 10.0 centimeters is 2.5N.
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What is Hooke's law?</h3>
If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.
F = kx
where k is the proportionality constant called the spring constant or force constant.
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.
Let the spring constant be very low 0.04N/m
The force applied is
F = 10 cm / 0.04
F = 0.1 m / 0.04
F = 2.5 N
Thus, the force result in stretching the spring 10cm is 2.5 N.
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Initial velocity of the object: 5 m/s
Explanation:
The figure in the problem is missing: find it in attachment.
The graph in the figure represents the velocity of an object (v) versus the time passed (t).
Here we are asked to find the initial velocity of the object.
This means that we have to find the velocity of the object when the time is zero, so when
t = 0
By looking at the corresponding value on the y-axis (velocity), we see that when t = 0, then
v = 5 m/s
Therefore, the initial velocity of the object is 5 m/s.
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