Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t = 
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
In Physics, displacement is the distance and direction of the shortest path
from the starting point to the ending point, regardless of the path followed
to get there.
The correct answer to this question is this one: B. 5.89 feet
<span>The area of a circular trampoline is 108.94 square feet. The radius of the trampoline is 5.89 feet.
</span>
For a circle,
A = (pi)r^2
108.94 ft^2 = (pi)r^2
r^2 = (108.94 ft^2)/(pi)
r = sqrt(108.94/pi) ft
r = 5. 89 ft
Answer:
490 nm
Explanation:
Wave length: This is the distance of a complete cycle covered by a wave.
The expression for frequency, wave length and speed is given as,
c = λf.................. Equation 1
Where c = speed of light in water, λ = wave length of light, f = frequency of light.
Also,
n = v/c............. Equation 2
Where n = refractive index of water, v = speed of sound in air.
Making c the subject of the equation,
c = v/n.............. Equation 3
Substitute equation 3 into equation 2
v/n = λf
make λ the subject of the equation
λ = v/nf............... Equation 4
Given: f = 4.6 × 10¹⁴ Hz
Constant: n = 1.33, v = 3.0×10⁸ m/s.
Substitute into equation 4
λ = 3.0×10⁸ /(1.33×4.6 × 10¹⁴ )
λ = 0.49×10⁻⁶ m
λ = 490 nm.
Hence the right option is 490 nm
Answer:
-15.3g
Explanation:
We can find the acceleration by using the following SUVAT equation:
where:
v = 0 is the final velocity (the headform comes to rest)
u = 6.0 m/s is the initial velocity before the impact
d = 12.0 cm = 0.12 m is the distance covered during the impact
a is the acceleration
Solving for a, we get
Where the negative sign means it is a deceleration.
Now we know that
So we can express the acceleration in units of g:
