Answer:
Explanation:
Given
mass of spring 
extension in spring 
downward velocity 
Position in undamped free vibration is given by

where 
also 



it is given


substituting values we get







Answer:
Magnitude of Vector = 79.3
Explanation:
When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:
x-component of vector = (Magnitude of Vector)(Cos θ)
y-component of vector = (Magnitude of Vector)(Sin θ)
where,
θ = angle of the vector with x-axis = 27°
Therefore, using the values in the equation of y-component, we get:
36 = (Magnitude of Vector)(Sin 27°)
Magnitude of Vector = 36/Sin 27°
<u>Magnitude of Vector = 79.3</u>
Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Answer:
B) Degrees
Explanation:
The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.
From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.
So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°