Hooks law: F=-x*K
F=Force
x=distance = 0.5m
K=constant
F=mass*9.81 = 30*9.81=aaaa [N]
aaa = -0.5 * K => K = something
Answer: Spring constant = 500 N/m
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No, the length doesn't affect the reading because the mass is constant and therefore, the weight is constant. The string's weight is so small that it can be neglected.
The stretch or compression of a spring is defined by Hooke's Law,where F is the Force acting on the spring,k is the constant spring factor, based on the material and construction of the spring.x is the displacement of the spring.The formula holds true until, or unless the force permanently deforms the spring material.
The k value is constant for the spring force project because the spring used is of the same material. As I recall this experiment with a wire, instead of a coiled spring, the wire had kinks in it -- so it was a compound spring. A kink would act as a spring until it was stretched out, and then it would add (a litttle) to the length of the wire. It was a mess to figure out. In your experiment, try adding weights of (x), (x+delta), (x+2*detla), where x is a weight and delta is a tiny, tiny weight. The k value might be constant over that small range of values.
Advantages: lighter, more compact, and easier to use. Weighing capacity tends to be bigger. Disadvantages: with repeated use, the spring in the scale can be permanently stretched. This tends to bring less accurate weights.
The oscillation of a spring is the motion that the spring makes when disturbed. Imagine holding the end of a spring and hanging a weight to the other end. If you do not disturb the weight, it will stay in a static position. However, when you pull down on the weight and let go, the spring "oscillates" up and down. The spring could also be compressed and released, creating the same effect. The up and down motion, which has a specific velocity and period relating to the spring constant k, is oscillation.
The answer is ni99er because add me on xbox. El Kemosabe F you if you need the answer lol
If a 2.5N extends the spring for 0.1m, so a 7.5N will extend 3 times that, a 0.3m.
Scales used to measure weight include: * Spring scale - measures the increased length of a spring as it stretches * Balance scale - uses a horizontal lever to compare unknown weights to that of known weights
No, the length doesn't affect the reading because the mass is constant and therefore, the weight is constant. The string's weight is so small that it can be neglected.
see me as earth
A spring scale measures weight. It is dependent on the gravity, and so one's weight in space, or on the moon would be different based on the spring scale. A balance measures mass, and since the standards in a balance are the same no matter what the gravity is, then the mass remains constant. However, for all practical purposes, the gravity on earth is constant so the weight and mass of the objects would be the same.
A spring scale measures weight. It is dependent on the gravity, and so one's weight in space, or on the moon would be different based on the spring scale. A balance measures mass, and since the standards in a balance are the same no matter what the gravity is, then the mass remains constant. However, for all practical purposes, the gravity on earth is constant so the weight and mass of the objects would be the same.
The net force acting in the stretch direction on the spring is proportional to its deformation (e.g., its stretch). In math talk that's F = kdX where k = 50 N/m and F = mg = 40*9.81 is the weight of the m mass. NOTE: 40 kg is not...not...a weight, it's a measure of mass. To get the weight, the force of gravity, we must multiply by g = 9.81 m/sec^2 which is a typical average acceleration due to gravity on the Earth's surface. So the stretch dX = F/k = mg/k = 40*9.81/50 = 7.848 meters. ANS.
If actually weighing the plates is impractical, you could try hanging the plates from a spring, and testing to find the spring's k value, and recording the displacement of the object while hanging from the spring, and use that to calculate the force on the plate, which equals mg. if the density is known, you could immerse the plates in something to find their volume and then calculate their weight from that. or, you could try and pull them with a force meter, taking two data points so that you can solve for the both the friction coefficient and weight.
The stretch or compression of a spring is defined by Hooke's Law,where F is the Force acting on the spring,k is the constant spring factor, based on the material and construction of the spring.x is the displacement of the spring.The formula holds true until, or unless the force permanently deforms the spring material.
The k value is constant for the spring force project because the spring used is of the same material. As I recall this experiment with a wire, instead of a coiled spring, the wire had kinks in it -- so it was a compound spring. A kink would act as a spring until it was stretched out, and then it would add (a litttle) to the length of the wire. It was a mess to figure out. In your experiment, try adding weights of (x), (x+delta), (x+2*detla), where x is a weight and delta is a tiny, tiny weight. The k value might be constant over that small range of values.