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Firdavs [7]
2 years ago
11

What property of equality can be used to solve the equation. n – 14 = 25

Physics
1 answer:
Inga [223]2 years ago
7 0

Do 25-14 and you will get your answer

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Air pressure is the wi get of air molecules pressing down on the earth. The pressure of the air molecules changes as you move upward from sea level into the atmosphere, the highest pressure is at sea level where the density of the air molecules is the greatest.
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Look at the illustration , what concepts of heat are being shown in this pictures <br>​
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2 years ago
A ball is thrown straight up. What will
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velocity at the top: 0 m/s

acceleration at the top: -9.8 m/s²

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Assuming up is positive and down is negative;

The velocity of the ball at the top of its path will be 0 m/s and the acceleration will be negative.

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The acceleration, however, is always negative no matter where the ball is in its motion. This negative acceleration causes the ball to slow down as it reaches the top, and speed up as it reaches the bottom.

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a cruise ship travels directly toward the dock with a velocity of 15 m/s relative to the water. A passenger walks 3 m/s in the s
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4 0
2 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
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