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Lynna [10]
2 years ago
12

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

s. The second piece has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s. What is the total momentum of the two pieces?
Physics
1 answer:
larisa [96]2 years ago
4 0

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, m_1=200\ g= 0.2\ kg

Mass of the second piece, m_2=300\ g= 0.3\ kg

Speed of the first piece, v_1=82\ m/s (along x axis)

Speed of the second piece, v_2=45\ m/s (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

p_x=m_1v_1

p_x=0.2\ kg\times 82\ m/s

p_x=16.4\ kg-m/s

p_y=m_2v_2

p_y=0.3\ kg\times 45\ m/s

p_y=13.5\ kg-m/s

The net momentum is given by :

p=\sqrt{p_x^2+p_y^2}

p=\sqrt{16.4^2+13.5^2}

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

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Therefore, it is a set of topics of mathematics that are relevent to introductory physics.

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Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.
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Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
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Answer:

Testing

Explanation:

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The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

4 0
2 years ago
A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
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Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

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Answer:

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