Answer:
The velocity of the arrow after 3 seconds is 30.02 m/s.
Explanation:
It is given that,
An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

We know that the rate of change of displacement is equal to the velocity of an object.

Velocity of the arrow after 3 seconds will be :

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =

0 - p × Area for trachea =



⇒

ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s
A pull between two objects, for example, between an object and Earth. When forces on an object are balanced, there is no change in speed or direction.
So the answer is I agree
Answer:
89 m
Explanation:
Applying
v = 2d/t................... Equation 1
Where v = velocity of sound in air, d = distance of the wall from Karen, t = time taken to hear the echo.
make d the subject of equation 1
d = vt/2..................... Equation 2
From the question,
Given: v = 343 m/s, t = 0.519 s
Substitute these values into equation 2
d = (343×0.519)/2
d = 89.01 m.
d ≈ 89 m