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sveticcg [70]
3 years ago
13

It is a general exercise that we do before practicing sports activities worm-up or cool down

Physics
2 answers:
ludmilkaskok [199]3 years ago
8 0

Answer:

We do warm-ups before we exercise. We do cool-downs after we exercise.

Explanation:

Warming up can help you get ready to do the actual exercise, so you are prepared.

Montano1993 [528]3 years ago
3 0

It is called a warm-up.

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Option C is the correct answer
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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
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Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
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Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude and θ.

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\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )

\vec{D}=  ( 2.121 m , -2.121 m )

and

\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )

\vec{E}= ( 2.71 m , 3.59 m )

Now, we can take the sum of the vectors

\vec{R} = \vec{D} + \vec{E}

\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )

\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m )

\vec{R} = ( 4.831 \ m , 1.469 \ m )

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}

|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}

|\vec{R}| = \sqrt{25.496 m^2}

|\vec{R}| = 5.049 m

To find the direction, we can use

\theta = arctan(\frac{R_y}{R_x})

\theta = arctan(\frac{1.469 \ m}{4.831 \ m})

\theta = arctan(0.304)

\theta = 16\°54'33''

As we are in the first quadrant, this is relative to the x axis.

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