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Inga [223]
3 years ago
7

a teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of t

he target each time. the drawing below shows where her arrows hit the target.
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
7 0
A teacher uses a bow and arrow to demonstrate accuracy and precision. she shoots several arrows, aiming at the exact center of the target each time. thedrawing below shows where her arrows hit the target.
<span> The statement  that best describes her shots is
</span><span>Her shots were neither accurate nor precise
</span>hope it helps
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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig
kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, \Delta S_{env} = -1.18 J/K

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

Learn more: brainly.com/question/22655760

6 0
2 years ago
What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when v
ad-work [718]

Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is \lambda.

Formula to calculate thickness of a benzene thin film is as follows.

     Optical path length through benzene (\lambda) = 2 \times n \times d

Hence, substituting the given values into the above formula as follows.

    Optical path length through benzene = 2 \times n \times d

                   d = \frac{\text{Optical path length through benzene}}{2 \times n}

                       = \frac{\lambda}{2 \times n}  

                       = \frac{615 \times 10^{-9}}{2 \times 1.501}   (as 1 nm = 10^{-9}m

                       = 204.9 m          

Thus, we can conclude that minimum thickness of benzene is 204.9 m.

4 0
3 years ago
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