Hi there!
We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).
Remember to break the velocity into its vertical and horizontal components.
Thus:
0 = vi - at
0 = 16sin(33°) - 9.8(t)
9.8t = 16sin(33°)
t = .889 sec
Find the max height by plugging this time into the equation:
Δd = vit + 1/2at²
Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²
Solve:
Δd = 7.747 - 3.873 = 3.8744 m
Answer:
IMA = 2.5 metres
EFFICIENCY = 80%
Explanation:
The AMA of a machine is referred to as the Actual Mechanical Advantage of a machine, calculated as the ratio of the output to the input force.
The Ideal Mechanical Advantage is the ratio of the input distance to the output distance.
From the diagram, the input distance which is also the distance moved by effort = 5metres
The load distance (output distance) = 2 metres
IMA = INPUT DISTANCE / OUTPUT DISTANCE
IMA = 5metres / 2 metres = 2.5 meters
Efficiency is the ratio of AMA TO IMA
AMA = 2, IMA = 2.5
EFFICIENCY = AMA / IMA
EFFICIENCY = (2 / 2.5) × 100%= 0.8 × 100%
EFFICIENCY = 80%
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 ×
m
wavelength λ = 700 nm = 700 ×
m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = 
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 × 
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 × 
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
Answer: The degree of the first term.
Explanation:
The function:

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.
Let the left hand side be donated by -x.
Then,

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 