It would be 1. B 2. A 3. A
Slowly; Boiling Point; Decrease; Decrease; Vibrate in place.
As temperature drops, so does thermal energy, and particle motion drops. The same trends in temperature, thermal energy, and motion applys to phases in decreasing order: gas>liquid>solid. The particle motion is always vibrations in place for solids because they are very tightly packed compared to liquids and gases.
The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
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To solve this problem we will use the concepts related to hydrostatic pressure. Which determines the pressure of a body at a given depth of a liquid.
Mathematically this can be described as

Here
= Density
g = Gravity
h = Height (Depth)
If we replace the values given in the equation we will have to


Therefore the pressure at the bottom will be 9.8kPa
Answer:
3.88 × 10^-4 m
Explanation:
Given that a person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 9.00 x 10^5 Pa. What is the area of contact between each tire and the ground?
The total weight = 614 + 85
The total weight = 699N
Let the total area of contact = A
Pressure = Force / A
Substitute all the parameters into the formula
900000 = 699 /A
A = 699 / 900000
A = 7.77 × 10^-4 m
The area of contact between each tire and the ground will be = A/2
That is, 7.77 / 2 = 3.88 × 10^-4 m
Therefore, the area of contact between each tire and the ground is 3.88 × 10^-4 m approximately.