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snow_tiger [21]
3 years ago
11

Which of the following is the least important factor of a personal fitness program? A. the individual's personal conditions B. t

he availability of resources C. the level of motivation D. the time of day physical activity will be performed Please select the best answer from the choices provided. A B C D Mark this and return
Physics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

I think it's B

Explanation:

I think its trying to tell you that no matter who you are you could still do regular fitness but I don't know‍♀️

You might be interested in
A +26.3 uC charge qy is repelled by a force
Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0
3 years ago
Suppose you now take the ball and using a bat, pop it straight up into the air with a hang-time of 5.00 s (the hang time is how
dolphi86 [110]

Answer:

<h3>30.66m</h3>

Explanation:

Using the equation of motion formula S = ut + \frac{1}{2}gt^2 where;

S is the height to which the ball rises

u is the initial velocity of the ball = 0m/s

a is the acceleration due to gravity = 9.81m/s²

t is the time taken by the ball in air = 5.0s

Note that the  time to rise to the peak is one-half the total hang-time = 5.0/2 = 2.5s

Substituting the given parameters into the formula above to get S:

S = ut + \frac{1}{2}gt^2\\\\S = 0(2.5)+ \frac{1}{2}(9.81)(2.5)^2\\\\S = 0+\frac{1}{2}(9.81)\times 6.25 \\\\S = \frac{61.3125}{2}\\ \\S = 30.65625m\\\\S \approx 30.66m

This means that the ball rises 30.66m before it reaches its peak.

8 0
3 years ago
A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?
eduard
First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock,  then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

       √(1 - v²/c²) .

You said that  v/c  = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² =  1 - 0.7225 = 0.2775

√(1 - v²/c²)  =  √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 =  <em>18.983 years </em>


I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

8 0
4 years ago
How long would it take a leopard, running at an average speed of 20 m/s to travel 500 m?
alexandr1967 [171]

Answer:

25 seconds

Explanation:

500/20

4 0
3 years ago
In the unsaturated zone, the pores of the soil are totally filled with water. T or F
creativ13 [48]







the answer is clearly false

6 0
3 years ago
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