-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s
-- The acceleration of gravity is 9.8 m/s².
-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.
-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
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-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
Answer:
An object may become extinct due to the lack of quantity of the object and the object no longer exists so it is now extinct.
Average Velocity = Total Displacement / Total time
1st part of journey, 350 km at velocity 125 km/h
Time = 350 / 125 = 2.8 hours.
2nd part of journey, 220 km at velocity 115 km/h
Time = 220 / 115 = 1.9 hours
Average Velocity = Total Displacement / Total time
= (350 + 220) / (2.8 + 1.9)
= 570 / 4.7 ≈ 121.3 km/hr
Average Velocity ≈ 121 km/hr due south.
Option C.
Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
Replacing the values we have:
Therefore the refractive index for the liquid is 1.7323
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
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