Explanation:
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Answer: 3.49 s
Explanation:
We can solve this problem with the following equation of motion:
(1)
Where:
is the final height of the ball
is the initial height of the ball
is the initial velocity (the ball was dropped)
is the acceleratio due gravity
is the time
Isolating :
(2)
(3)
Finally we find the time the ball is in the air:
(4)
Answer:
Shiny metals such as copper, silver, and gold are often used for decorative arts, jewelry, and coins.
Strong metals such as iron and metal alloys such as stainless steel are used to build structures, ships, and vehicles including cars, trains, and trucks.
Some metals have specific qualities that dictate their use. For example, copper is a good choice for wiring because it is particularly good at conducting electricity. Tungsten is used for the filaments of light bulbs because it glows white-hot without melting.
Nonmetals are plentiful and useful. These are among the most commonly used:
Oxygen, a gas, is absolutely essential to human life. Not only do we breathe it and use it for medical purposes, but we also use it as an important element in combustion.
Sulfur is valued for its medical properties and as an important ingredient in many chemical solutions. Sulfuric acid is an important tool for industry, used in batteries and manufacturing.
Chlorine is a powerful disinfectant. It is used to purify water for drinking and fill swimming pools.
Explanation:
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
Answer:
as a way to put two things together and yeeah
Explanation:
i know because this is what my mom told me
