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dangina [55]
3 years ago
15

Help me with number 2 please

Physics
1 answer:
NARA [144]3 years ago
3 0
The answer would be B. An example of this would be the Noble gasses. They include: Helium, Argon, Neon, and so on. They are all located on the very right side because they share similar chemical behaviours; they dont react very easily because they have a full valence shell.
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Static friction because the rocks arent moving are they?
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An object is thrown with velocity v from the edge of a cliff above level ground. Neglect air resistance. In order for the object
musickatia [10]

Answer:

(C) greater than zero but less than 45° above the horizontal

Explanation:

The range of a projectile is given by R = v²sin2θ/g.

For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°

2θ = 90°

θ = 90°/2 = 45°

So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.

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Define , gravitational acceleration
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3 years ago
A fish takes the bait and pulls on the line with a force of 2.3 N. The fishing reel, which rotates without friction, is a unifor
natulia [17]

Explanation:

The given data is as follows.

     Pulling force on the reel is F = T = 2.3 N

     Mass of cylinder (m) = 0.82 kg

    Radius of cylinder (r) = 0.045 m

Formula for torque pulling force on the cylinder is as follows.

           \tau = Fr

                     = I \times \alpha

Moment of inertia of the cylinder (I) is as follows.

            I = \frac{mr^{2}}{2}

  \alpha = angular acceleration of the cylinder

Hence,

                    Fr = (\frac{mr^{2}}{2}) \alpha

or,   \alpha = \frac{2F}{mr}

                   = \frac{2 \times 2.3 N}{0.82 kg \times 0.045 m}

                   = 124.66 rad/s^{2}

Amount of line pulled is h and it is in a times of 0.2 sec.

Now, linear acceleration is calculated as follows.

             a = r \alpha

                = 0.045 m \times 124.66 rad/s^{2}

                 = 5.61 m/s^{2}

Now, relation between h and acceleration as follows.

             h = v_{o}t + \frac{1}{2}at^{2}

here,   v_{o} = 0

Hence, calculate the value of h as follows.

             h = v_{o}t + \frac{1}{2}at^{2}

                = 0 + \frac{1}{2} \times 5.61 m/s^{2} \times (0.2s)^{2}

                = 0.1121 m

Thus, we can conclude that acceleration of the fishing reel is 5.61 m/s^{2} and it will pull 5.61 m/s^{2} line from the reel in 0.20 s.

7 0
4 years ago
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