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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The gravitational force between two objects may be calculated by the equation,
F = G x (m1 x m2) / d²
where G is universal gravitational constant, m1 and m2 are the masses of two bodies and d is the distance between them. Substituting the known values,
9.20 x 10^-9 = (6.673 x 10^-11) (5.01 kg x m2) / (0.192m)²
The value of m2 is approximately 1.01 kg.
Answer:
the DISTANCE between the lever arm and the force is always 90º
Explanation:
In this exercise, you are asked to complete the missing words so that the phrase makes sense.
note that the torque is
τ = F x r
where bold indicates vectors
When the rope is pulled, the DISTANCE between the lever arm and the force is always 90º
Answer:
the correct answer is the 60

Answer:
G. It will take twice as long.
Explanation:
Let's call
the original speed of the plane and
the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

In this problem, we are told that the plane encounters wind moving at half of its speed:
, in the opposite direction. This means that the new speed of the plane is

And so, the time the plane takes now to complete the flight is

So, the plane takes twice the time as before.