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pochemuha
1 year ago
10

A moving bomb explodes into two pieces. A 2 kg fragment moves to the right at 15 m/s and the other fragment of mass 10 kg moves

to the left at 6 m/s. What was the initial speed
Physics
1 answer:
Alla [95]1 year ago
8 0

The initial velocity is obtained as 2.5 m/s.

<h3>What is the initial speed?</h3>

To obtain the initial speed, we have to apply the law of conservation of linear momentum which states that momentum before collision is equal to momentum after collision.

Now, the two fragments initially had one velocity before they were split in two directions opposite each other;

Applying the principle of conservation of linear momentum to this problem;

(2 + 10)v = (6 * 10) - (2 * 15)  (they moved in opposite directions)

12v =60 - 30

12v = 30

v = 30/12

v = 2.5 m/s

Learn more about momentum:brainly.com/question/24030570

#SPJ1

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According to Newton's law of universal gravitation, which statement is true?
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Answer:

Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

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Ian’s school is exactly 6 blocks north of his house. What is his average velocity on a walk from home to school that takes him 1
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Answer:

.5 units north

<em><u>or</u></em>

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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
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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

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