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dexar [7]
2 years ago
13

A Nerf gun is shot straight up from the ground with an initial velocity of 14.0 m/s. What is the total time the projectile is in

the air?
PLEASE SHOW WORK WILL MARK BRAINLIEST
Physics
1 answer:
Sergeeva-Olga [200]2 years ago
7 0

The projectile has a height <em>h</em> at time <em>t</em> given by

<em>h</em> = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for <em>t</em> when <em>h</em> = 0 :

0 = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²

0 = 1/2 <em>t</em> (28.0 m/s - <em>g t</em> )

1/2 <em>t</em> = 0   <u>or</u>   28.0 m/s - <em>g</em> <em>t</em> = 0

The first equation says <em>t</em> = 0, which refers to the moment the gun is first fired, so we ignore that solution. We're left with

28.0 m/s - <em>g t</em> = 0

<em>t</em> = (28.0 m/s) / <em>g</em>

<em>t</em> = (28.0 m/s) / (9.80 m/s²)

<em>t</em> ≈ 2.86 s

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A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).
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Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles (\vec r_{cm}), measured in meters, is defined by this weighted average:

\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}} (1)

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m_{i} - Mass of the i-th particle, measured in kilograms.

\vec r_{i} - Location of the i-th particle with respect to origin, measured in meters.

If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})

(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

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b) The y coordinate of the third mass is -0.944 meters.

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Answer: 0.4 m

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With speed of ambulance being (61.9 m/s) -> We solve using

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