Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
<u>θ₀ = 5.22°</u>
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
<u>θ₀ = 84.78°</u>
Answer: Find the answer in the explanation
Explanation: Given the Roman numeral and the representation
I. part of a coal-fired power plant
II. part of a nuclear power plant
III. part of a coal-fired power plant and part of a nuclear power plant
a.) Boiler : I
b.) Combustion chamber: I
c.) Condenser: I
d.) Control rod: II
e.) Generator: III
f.) Turbine: III
Toward the end processes part of both coal fire and nuclear power, they both make use of turbine and generator to generate electricity.
We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,
With this function we should only calculate the derivate in function of c
That is the rate of change of 
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.
Hence we have finally the rate of change when c=0.
a. The disk starts at rest, so its angular displacement at time 
is
It rotates 44.5 rad in this time, so we have
b. Since acceleration is constant, the average angular velocity is
where 
is the angular velocity achieved after 6.00 s. The velocity of the disk at time is
so we have
making the average velocity
Another way to find the average velocity is to compute it directly via
c. We already found this using the first method in part (b),
d. We already know
so this is just a matter of plugging in 
. We get
Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that
Then for 
we would get the same 
.
This is the upthrust on an object which is placed inside a fluid
This force act upwards and always push upwards
so the correct answer is given as
D. A force within a fluid that pushes upward
this force is always due to pressure difference at two levels of
at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy
