Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
The answer is A. digestion of food. The chemical change to matter means that the molecule structure has been destroyed. For B, C and D. the matter are still ice, paper and ice. So they are physical change.
The correct answer is option #4. The greatest obstacle to developing solar energy is large areas required to collect sufficient energy. Solar energy is the energy that is the sun is producing and being converted into thermal energy that can be used for everyone's daily lives.
Answer:
The Solar System is the Sun and all the objects that orbit around it
The Sun is orbited by planets, asteroids, comets and other things. ...
It has strong gravity
There are many types of energies
three types are
kinetic energy
potential energy
gravitational energy
The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
- As we know that the expression of pressure as,
where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,
- From this, the value of weight will be,
Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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