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Reika [66]
3 years ago
5

Discuss the difference between short-term and long-term fitness goals. Provide an example of each

Physics
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

The difference between a short-term and a long-term fitness goal is that short-term fitness goals are ones you could complete in a short period of time, a long term goal is something you'd work on for months or years.

An example of a short-term fitness goal: working out 2 times a week

An example of a long-term fitness goal: having a better diet or more lifestyle improvements

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What type of lamp is used in a spectrophotometer to produce visible light?
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A ball is thrown directly downward with an initial speed of 8.50 m/s, from a height of 29.7 m. After what time interval does it
stellarik [79]
Using the following formula for linear-motion, the missing variable can be solved:

s = Vi * t + 1/2 (a * t^2) 

Where: s = displacement = 29.7 m
Vi = initial velocity = 8.5 m/s
a = acceleration = 9.8
t = time = ?

Substituting:

29.7 = 8.5t + 1/2 (9.8*t^2)
29.7 = 8.5t + 4.9t^2

Dividing both sides by 4.9:

6.06 = 1.73t + t^2
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From the above values, the correct answer is 1.74 seconds.
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A natural water molecule (H2O) in its vapor state has an electric dipole moment of magnitude, p = 6.2 x 10-30 C.m. (a) Find the
Vika [28.1K]

Answer:

a    D = 3.9 *10^{-12} \ m

b    \tau_{max} = 1.24 *10^{-25} \  N\cdot m

c   W =  2.48 *10^{-25} J

 

Explanation:

From the question we are told that

   The magnitude of electric dipole moment is  \sigma  =  6.2 *10^{-30} \ C \cdot m

     The electric field is E =  2*10^{4} \ N/C

   

The distance between the positive and negative charge center is mathematically evaluated as

     D =  \frac{\sigma }{10 e}

Where  e is the charge on one electron which has a constant value of  e = 1.60 *10^{-19} \ C

  Substituting values

     D =  \frac{6.20 *10^{-30}}{10 * (1.60 *10^{-19})}

      D = 3.9 *10^{-12} \ m

The maximum torque is mathematically represented as

       \tau_{max} = \sigma * E  * sin (\theta)

Here  \theta  =  90^o

This because at maximum the molecule is perpendicular to the field

    substituting values

       \tau_{max} =  6.2 *10^{-30} * 2*10^{4} sin ( 90)

       \tau_{max} = 1.24 *10^{-25} \  N\cdot m

The workdone is mathematically represented as

      W =  V_{(180)} - V_{0}

where   V_{(180)} is the potential energy at 180° which is mathematically evaluated as

     V_{(180) } = -   \sigma  * E  cos (180)

Where the negative signifies that it is acting against the  field

   substituting values

     V_{(180) } = -   6.20 *10^{-30}  * 2.0 *10^{4}  cos (180)

      V_{(180) } = 1.24*10^{-25} J

and

     V_{(0)} is the potential energy at 0° which is mathematically evaluated as

            V_{(0) } = -   \sigma  * E  cos (0)

   substituting values

     V_{(0) } = -   6.20 *10^{-30}  * 2.0 *10^{4}  cos (0)

      V_{(0) } =- 1.24*10^{-25} J

So W =  1.24 *10^{-25} - [-1.24 *10^{-25}]

    W =  2.48 *10^{-25} J

6 0
4 years ago
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