Recall this gas law:
= 
P₁ and P₂ are the initial and final pressures.
V₁ and V₂ are the initial and final volumes.
T₁ and T₂ are the initial and final temperatures.
Given values:
P₁ = 475kPa
V₁ = 4m³, V₂ = 6.5m³
T₁ = 290K, T₂ = 277K
Substitute the terms in the equation with the given values and solve for Pf:

<h3>P₂ = 279.2kPa</h3>
<h2>Answer: about the same size of the gap or slit</h2>
Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.
Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.
In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.
Therefore:
<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap
</u></h2>
<u />
The correct option is SCALE.
This is because the experimental data you are interested in is mass and scale is the instrument that is used to measure mass. To use the scale, measure the mass of your water before you put it inside the sun and measure the mass again after you remove it from the sun. The difference in mass is the quantity of water that is evaporated.
There are different type of scale and the one that you use will depend on the size of the material that you want to measure.<span />
Answer:
creamy with milk and chocolate
Answer:d
Explanation:
Suppose V is the voltage of battery and R is the resistance of bulb
so Power drop for initial stage

When another identical bulb of same resistance is applied in parallel so voltage Drop across both the resistor will be same i.e. V
so Power consumed 
so there is no change in power and hence no dip in brightness