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Angelina_Jolie [31]
2 years ago
7

A bullet having the mass of 110g is fired. If it's velocity is 50m/s, calculate the kinetic energy​

Physics
2 answers:
Papessa [141]2 years ago
7 0

\text{Kinetic energy,}\\\\E_k = \dfrac  12 mv^2\\\\~~~~~=\dfrac 12 \cdot 110  \times 10^{-3} \cdot 50^2\\\\~~~~~=137.5~ \text{J}

natulia [17]2 years ago
4 0

Answer: 137.5 Joules

Explanation:

The SI units for kinetic energy is Joules, the SI units for mass is kilograms, and the SI units for velocity is meters per second. We are given the mass in grams, so we must convert it to kilograms. There are 1,000 grams in 1 kilogram:

110g*\frac{1kg}{1000g} =0.11kg

Kinetic energy is the energy the system has due to its motion. It is defined to be:

KE=\frac{1}{2} mv^2

Plug everything in:

KE=\frac{1}{2} (0.11kg)(50m/s)^2

KE=137.5J

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
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Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

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First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

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\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

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c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

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