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Eduardwww [97]
4 years ago
5

A person needs to be ready both __________ and __________ before starting an exercise plan.

Physics
2 answers:
kati45 [8]4 years ago
4 0
I think it is mentally and physically 
Harrizon [31]4 years ago
4 0
Mentally and physically
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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
Eva8 [605]

Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
  • Direction of R relative to the x axis= 16°54'33'

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.

So, for our vectors, we will have:

\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.

3 0
4 years ago
What is the SI unit used to mesasure force
lukranit [14]

Answer:

The si unit to measure force are Pascels or Pa

Hope this helps

Please mark me as Brainliest

6 0
3 years ago
A __________ psychologist helps children with learning problems and works with teachers and parents to create healthy learning e
Natali [406]
C seems the most likely.
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4 years ago
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A 0.280 m radius, 475 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
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Hope it helps u mate. Plz follow me .

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4 years ago
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal
kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both  spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

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3 years ago
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