Blood flows through the major artery at 1 m/s for 0.5 m then at a 0.6 m/s over a distance of another 0.5 m through the small art
1 answer:
Blood flows through the major artery at 1 m/s for 0.5 m then at a 0.6 m/s over a distance of another 0.5 m through the small artery the average speed of blood is 0.4 m/s.
We know that average speed = =0.4 m/s
Average speed is an important component in determining how long it takes to finish a journey. Average speed is simply a technique that assists us in calculating trip time and distance. It is obvious that the speed changes throughout the travel, making determining the average speed even more critical.
There are various methods for determining an object's or vehicle's average speed.
It is most desired when the speed of the object remains constant during the voyage, i.e. does not rise or decrease.
The approach for determining the average Speed is to divide. Divide the distance the vehicle travels by the time it travels to get the result.
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Answer:
Explanation:
The formula for force is:
If we rearrange the formula to solve for a (acceleration), the formula becomes
The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².
The mass is 2 kilograms.
Substitute the values into the formula.
Divide. Note that the kilograms will cancel each other out (hence why we changed the units).
The acceleration is<u> </u><u>34 meters per second squared.</u>
Answer: a) The cliff is 532.05m high
b) Her speed just before hitting the ground is 102.12 m/s
Explanation: To solve This, I'll use a sketch diagram, attached to this solution,
In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,
Y1 = ut + 0.5g(t^2)
Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s
Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m
Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.
Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,
Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds
Y2 = 340 × 3 = 1020m
But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x
x = 487.95m
So, total height of cliff = 44.1 + 487.95 = 532.05m
b) the speed of the teacher just before she hits the ground.
Using the equations of motion again,
(V^2) = (U^2) + 2gs
Where v is the final velocity to be calculated
U is the initial velocity = 0m/s
g is acceleration due to gravity = 9.8m/s2
S is the total height she fell through, that is, the height of the cliff = 532.05m
(V^2) = 0 + 2×9.8×532.05 = 10428.18
V = √(10428.18) = 102.12m/s
QED!
