Answer:
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water the weight of the two beakers is the same
Explanation:
The beaker weight is
beaker A
W_total = W_ empty + W_water
Beaker B
W_total = W_ empty + W_water + W_roca
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same
Explanation:
Precision represents that how close the different measurements of the sample one take are to one another.
- One can increase the precision in lab by paying attention to each and every detail.
- Usage of the equipment properly and also increasing the sample size.
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Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
- Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.
Answer:
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Sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation. In an auditorium or big hall excessive reverberation is highly undesirable
Explanation:
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Answer:
When an electromagnetic wave passes from space to matter, some part of the energy is absorbed by the matter and it increases its energy. The wave may reflect and some part may pass through the matter depending on the amount of energy they have. The amplitude of the wave decreases if some parts of it are reflected.
Answer:

Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:

where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:
