Answer:
Explanation:
Using the pythagoras theorem, the displacement is expressed as;
d² = x²+y²
y = 36m (north)
x = 20m east
Substitute;
d² = 36²+20²
d² = 1296+400
d² = 1696
d = √1696
d = 41.18m
For the direction;
theta = tan^-1(y/x)
theta = tan^-1(36/20)
theta = tan^-1(1.8)
theta = 60.95°
Hence the magnitude is 41.18m and the direction is 60.95°
A process known as fixation<span>. the majority of nitrogen is fixed by </span>bacteria<span>, most of which are </span>symbiotic<span> with plants</span>
<span />
Option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>
<u></u>
<h3>What is an arch?</h3>
An arch is indeed a vertical curving construction that covers an elevated space that may or may not sustain the load above it or the pressure gradient against it
In the case of a horizontally arched, such as an embankment dam. While arches and vaults are often confused, A vault is defined as an ongoing arch forming a roof.
Option D satisfies the fill-in blanks option.
Hence option D is correct. An arch carries the thrust of weight to its <u>sides </u>with a <u>post-and-lintel.</u>
<u></u>
To learn more about the arch refer to the link;
brainly.com/question/18162421
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
The x- and y-coordinates are 9142.57 m and -304.425 m
<u>Explanation:</u>
As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell
in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of
and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by


For motion with constant acceleration, we know

Along the horizontal, x-axis, we might write this as

Measuring distances relative to the firing point means

we know that,

or,

By applying the values, we get,

The acceleration of gravity is vertically downward and is
, hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

we know,
and
, so,

y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m