Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
The time taken for the tiny saliva to travel is 0.55 second.
The horizontal distance traveled at speed of 4 m/s is 2.2 m.
The horizontal distance traveled at speed of 20 m/s is 11 m.
<h3>
Time of motion of the tiny saliva</h3>
The time taken for the tiny saliva to travel is calculated as follows;
h = vt + ¹/₂gt²
where;
- v is initial vertical velocity = 0
- g is the acceleration due to gravity
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
Substitute the given parameters and solve for time of motion;
t = √(2 x 1.5 / 10)
t = 0.55 second
<h3>Horizontal distance traveled at speed of 4 m/s</h3>
X = Vx(t)
X = (4 m/s)(0.55)
X = 2.2 m
<h3>Horizontal distance traveled at speed of 20 m/s</h3>
X = (20)(0.55)
X = 11 m
Learn more about time of motion here: brainly.com/question/2364404
#SPJ1
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.
The equation that describes the beat frequency is
For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,
A) The frequency of the violin would be given by
B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.
