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ladessa [460]
3 years ago
13

The results of a dart game were precise but not accurate. The accepted value of the game was the center of the dartboard. Which

correctly describes the results? All of the darts hit the center of the board. Some of the darts hit the center of the board. The darts hit the same general area of the board. The darts hit very different areas of the board.
Physics
2 answers:
hoa [83]3 years ago
8 0

Answer:

The darts hit very different areas of the board.

Explanation:

Pavel [41]3 years ago
3 0

Answer:

The darts hit the same general area of the board.

Explanation:

The meaning of precision is the closeness of measured values to other measured values. Therefore, since the results of the dart game were precise and inaccurate, the darts would hit the same general area because the measured values should be close to one another.

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ASAP!! please What is the answer? MCQ. Thank you
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Explanation:

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"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel
adelina 88 [10]

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \sqrt{2g y_o}

         v = \sqrt{2 \ 9.8\ H}

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

          v = w r

          w = v / r

the momentorot of inertia indicates that it is worth

          I = \frac{2}{5} m r²

we substitute

           m g H = ½ m v² + ½ (\frac{2}{5}  m r²) (\frac{v}{r} )²

           gH = \frac{1}{2}  v² + \frac{1}{5}  v² = \frac{7}{10}  v²

           v = \sqrt{\frac{10}{7} \ g H}

           v = \sqrt{ \frac{10}{7}  \ 9.8 \ H}

           v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

           sin θ = H / L

           L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

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" ... product of the masses ... "
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Answer:

230.26 N

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7 0
3 years ago
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