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  1. Last 7 days
  2. math.libretexts.org math.libretexts.org
    1.3: Using Letters to Symbolize Assertions
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    1. sky1381 15 Sep 2026
      In the remainder of this , we will discuss a logical language called . It provides a convenient way to describe the logical relationship between two (or more) assertions, by using capital letters to represent assertions. Considered only as a symbol of , the letter A could mean any assertion. So, when translating from English into , it is important to provide a symbolization key that specifies what assertion is represented by each letter.

      The name of the logical language and the symbol type A is supposed to be are never specified.

      Based upon 1.4, the name is probably supposed to be 'propositional logic.'

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    math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/01:_Propositional_Logic/1.03:_Using_letters_to_symbolize_assertions
  3. Sep 2026
  4. phys.libretexts.org phys.libretexts.org
    6.5: Friction (Part 2)
    1
    1. sky1381 06 Sep 2026
      A skier with a mass of 62 kg is sliding down a snowy slope at a constant velocity.

      How can the skier be at a constant velocity? The force of friction is stated to only be 45 N, but the component of the gravitational force that is parallel to the surface of an object on a 25 deg inclined plane is m*g*sin 25=-257 N. However, that would result in an acceleration of 3.4 m/s^2 in the negative-x direction. So either, the skier must not be at a constant velocity, or the skier must have an extra force acting on them, which would result in the FBD in the Strategy Section being incorrect.

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    phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book:_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/06:_Applications_of_Newton's_Laws/6.05:_Friction_(Part_2)
  5. Aug 2026
  6. phys.libretexts.org phys.libretexts.org
    5.E: Newton's Laws of Motion (Exercises)
    1
    1. sky1381 20 Aug 2026
      Shown below is a body of mass 1.0 kg under the influence of the forces F→A, F→B, and mg→. If the body accelerates to the left at 20 m/s2, what are F→A and F→B?

      The high acceleration in the left direction results in F_A having a negative magnitude (or the body must have some upward acceleration).

      Assuming F_A is zero, then F_B must have a x-component of 20 N to produce an acceleration of 20 m/s^2. However, as F_B is angled at 30 degrees, it must also produce a vertical force of 10 N, which is greater than the weight of the body (-9.8 N). This causes the body to move vertically upward. Having F_A be any value above zero causes this problem to get worse. They only way to allow the body to accelerate at 20 m/s^2 in x while having no acceleration in y is to have F_A be in the opposite direction.

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    phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book:_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/05:_Newton's_Laws_of_Motion/5.E:_Newton's_Laws_of_Motion_(Exercises)
  7. phys.libretexts.org phys.libretexts.org
    4.4: Projectile Motion
    1
    1. sky1381 07 Aug 2026
      53.1° below the horizontal at the point of impact.

      Shouldn't this be -36.9 degrees below the horizontal? The inverse tangent shown in 4.4.32 has the x-velocity on top of the y-velocity, while the inverse tangent should actually have the y-velocity on top.

      Also, it wouldn't be possible for the velocity to be 53.1 degrees below the horizontal before dropping below the point it was released, as the ball left at 45 degrees above the horizontal, and would reach the same y-position as it had before it left with the same angle below the horizontal, so the ball would have to fall below the point it was released to reach 53.1 degrees below the horizontal.

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    phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book:_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/04:_Motion_in_Two_and_Three_Dimensions/4.04:_Projectile_Motion
  8. phys.libretexts.org phys.libretexts.org
    4: Unit Conversion
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    phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book:_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/01:_Units_and_Measurement/04:_Unit_Conversion
  9. phys.libretexts.org phys.libretexts.org
    7.3: Kinetic Energy
    1
    1. sky1381 04 Aug 2026
      Solution Do not forget to convert km into m to do these calculations, although, to save space, we omitted showing these conversions. K=12⁢(80k⁢g)⁢(10m/s)2=4.0k⁢J. m=2⁢Kv2=2⁢(4.2×1023J)22k⁢m/s)2=1.7×1015k⁢g. K=12⁢(1.68×110−27k⁢g)⁢(2.2k⁢m/s)2=4.1×10−21J. Significance

      There's an extra 1 on the (1.68110^-27 kg) tern, it should be (1.6810^-27 kg)

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    phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book:_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/07:_Work_and_Kinetic_Energy/7.03:_Kinetic_Energy