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makvit [3.9K]
3 years ago
7

Arm

Physics
1 answer:
Daniel [21]3 years ago
3 0
Can't say for sure but I think it's a) or c)
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In his psychology class, Professor Adams said that fear of spiders and snakes are common in humans, but fear of automobiles is n
Svetradugi [14.3K]

Answer:

Evolutionary

Explanation:

Because our ancestors were afraid of these poisonous spiders and snakes which lead to them to telling their kids of these dangers. Their kids will fear them without even encountering any and this will also be passed on to their future generations.

5 0
3 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
2. Can the frictional force in this experiment be ignored? Explain.
neonofarm [45]

Answer:

i dont know

Explanation:

i dont know

5 0
2 years ago
Pa help po science po yan pang grade 7
AveGali [126]

Answer:

table 1:

1. 100/15 = 6.7 m/s

2. 100/12 = 8.3 m/s

3. 100/9 = 11.1 m/s

table 2:

1. 100/8 = 12.5 m/s

2. 100/6 = 16.7 m/s

3. 100/4 = 25 m/s

Explanation:

8 0
3 years ago
1) Archeologists have found fossil remains of some of the first land
dimaraw [331]

Answer: Carbon 14 and Uranium 238 are not used together to determine fossil ages.

Explanation:

Carbon 14 with a half life of 5,700 years can only be used to date fossils of approximately 50,000 years. Most fossils are thought to be much older than 50,000 years. Also most fossils no longer contain any Carbon. The fossilized remains have been mineralized where the original organic material has been replaced and turned into stones containing no carbon.

Uranium 238 has a half life of 4.5 billion years. Uranium can be used to date the age of the earth. If 50% of pure uranium' is left in a sample the sample is assumed to be 4.5 billion years old.( This is assuming that the original sample was 100% uranium and no Uranium 238 has been eroded or lost in 4.5 billion years old. If a fossil has only 25 % of the Uranium 238 the sample has an estimated age of 3.2 Billion years. This would be the estimated age of the earliest life or formation of fossils.

Note no fossils contain Uranium 238. Uranium 238 is only found in igneous or volcanic rocks. So no fossils can be dated directly using U 238.

Because of the huge differences in the half lives of Carbon 14 and Uranium238 they cannot be used together. Carbon 14 can only be used to date fossils of a very recent age. Uranium 238 can only be used to date volcanic rocks of a very old age.

3 0
3 years ago
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