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Alexus [3.1K]
2 years ago
11

How far (in meters) will you travel in 3 minutes running at a rate of 6 m/s?

Physics
1 answer:
trasher [3.6K]2 years ago
7 0
The formula for distance is speed times time.

So 3 times 6 would be 18 m/s
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lightning and thunder both occur at the same time why do you see the Lightning before you hear the thunder
denis23 [38]

Because the light from it travels to you about 874 thousand times
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Look at the rock sitting on the hill in the picture above. Gravity should make the rock slide down the hill. What force is actin
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A car traveling at 50 m/s comes to a stop in 5 seconds. What is the<br> acceleration of this car?
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4 0
2 years ago
A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be ______
snow_tiger [21]

Answer:

The body must be moving with a constant non zero acceleration.

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4 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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