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Sauron [17]
3 years ago
12

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

he second 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 ?
a) 361 kg x m/s at 56.3 degrees from the x-axis .
b) 93.5 kg x m/s at 28.8 degrees from the x-axis .
c) 21.2 kg x m/s at 39.5 degrees from the x-axis .
d) 361 kg d m/s at 0.983 degrees from the x-axis .
Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis  ( tan^-1 (13.5 / 16.4)

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Answer: A. The total displacement divided by the time and  C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity V will be expressed by the following equation:

V=\frac{d}{t}

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On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

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3 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
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A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
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hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
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1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
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