Answer:
the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Explanation:
Given;
initial velocity of the airplane. u = 34.5 m/s
distance traveled by the airplane, s = 46,100 m
final velocity of the airplane, v = 40.7 m/s
The acceleration of the airplane is calculated from the following kinematic equation;
v² = u² + 2as

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11
Answer:
Answer:This organism may be identified by its color, the spines on its back, the antennae, and therefore the long, thin body. There are many other characteristics that might even be wont to identify this organism.
Explanation:
Refer to the diagram shown below.
From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°
From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.
The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.
Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.