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julsineya [31]
3 years ago
13

Six parts of Health-Related Fitness

Physics
1 answer:
mars1129 [50]3 years ago
8 0
Considering the total body, there are six elements of fitness: aerobic capacity, body structure, body composition, balance, muscular flexibility and strength
You might be interested in
Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
Masja [62]

Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

6 0
3 years ago
Two resistors of 12Ω and 2Ω are connected in parallel to a battery of 4 V. Current through 12Ω resistor will be _____________ an
Alex

Answer:

0.33  Amp

8 Watts

Explanation:

The current through the 12 Ω resistor will be given by Ohm's law:

4 V / 12 Ω = 1/3 Amp ≈ 0.33  Amp

The power dissipated on the 2 Ω resistor can be calculated via the formula: \frac{V^2}{R}  which gives:

4^2 / 2 = 8 Watts

6 0
3 years ago
It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?
NeX [460]

Answer:

A (2066,6 N)

Explanation:

Use the Work formula

62.000J = F . 30

62.000/30 = 2066,6 N

The amout of time it took to move the rock doesn´t matter at all.

It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.

5 0
3 years ago
100 points!!! Please help!!!
Greeley [361]

Answer:

41°

Explanation:

Kinetic energy at bottom = potential energy at top

½ mv² = mgh

½ v² = gh

h = v²/(2g)

h = (2.4 m/s)² / (2 × 9.8 m/s²)

h = 0.294 m

The pendulum rises to a height of  above the bottom.  To determine the angle, we need to use trigonometry (see attached diagram).

L − h = L cos θ

cos θ = (L − h) / L

cos θ = (1.2 − 0.294) / 1.2

θ = 41.0°

Rounded to two significant figures, the pendulum makes a maximum angle of 41° with the vertical.

7 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
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