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Leno4ka [110]
2 years ago
10

In the middle of your Fitness drug you start to feel a little weak selective method of monitoring your Fitness level which would

be readily available to you at the least cost: A. Heart rate monitor B. Pedometer C. Talk test D. Stopwatch
Physics
2 answers:
salantis [7]2 years ago
8 0

Answer:

c

Explanation:

Eva8 [605]2 years ago
7 0

Answer:

C. Talk test

Explanation:

The talk test would be readily available to me at the the least cost.

The talk test is about the easiest way that one can monitor intensity as they exercise. Because the only thing needed here is the ability to talk and to breathe.

The intensity lies on if one can talk and breathe at the same time. The harder one exercises, the more breathless they become and they find it difficult to talk.

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Please answer as soon as possible. <br><br> A Physics question about electricity and circuits.
Licemer1 [7]

Answer:

Explanation:

The formula P=VI, just rearrange to find I

So

I=P/V

=75/120

=0.625amps

625mA

3 0
3 years ago
The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th
Yuri [45]

Answer:

R=64.32\ lb\\\\\theta=84.3\°

Explanation:

Given:

Ratio of lift force to drag force is, \frac{L}{D}=10

Lift force on a short section is, L=64\ lb

Magnitude of resultant, R= ?

The angle of 'R' with the horizontal is, \theta=?

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb

Now, the magnitude of resultant 'R' is given as:

R=\sqrt{L^2+D^2}

Plug in the given values and solve for 'R'. This gives,

R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle \theta is given as the arctan of the ratio of the lift and drag force.

Therefore,

\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°

Therefore, the angle made with the horizontal is 84.3°.

6 0
3 years ago
Por que no céu sem nuvens a lua nova não é visivel e a lua cheia aparece em grande destaque?
weeeeeb [17]
Porque la matematica es imposible

4 0
3 years ago
A power plant uses Uranium<br> to produce energy
antoniya [11.8K]

Answer:

how is that a question?

Explanation:

yeah i dunno the answer cause thats not a question

7 0
3 years ago
I’m not sure how to solve this
spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

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

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0
2 years ago
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