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maw [93]
4 years ago
15

Consider the differences between anaerobic and aerobic exercises. Is one of these exercises more important than the other in an

exercise program? Support your answer with examples.
Physics
2 answers:
kodGreya [7K]4 years ago
6 0
Anaerobic exercise is exercise that doesn’t involve the use of Oxygen whereas Aerobic exercise is with Oxygen. It depends on the sport you are participating in when doing an exercise program but sprinters are better suited to anaerobic exercises because that is how their races would be, because they are running in short bursts so wouldn’t have the time to use a lot of Oxygen whereas long-distance runners would use an exercise program that focusses on aerobic exercises because they’re running for a long time so would constantly be needing Oxygen.
kicyunya [14]4 years ago
3 0

A sample response follows: Both types of exercises are important for a person's health. Anaerobic exercises build muscle mass and strength, which is important in maintaining good health. Aerobic exercise build endurance and cardiorespiratory fitness.

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If you travel 5 miles north then turn and travel 5 miles south, you are now _____ miles from where you started.
horsena [70]
0 miles from where you started
3 0
3 years ago
Read 2 more answers
A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t
LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

F_{c} = \frac{mv^{2}}{R}  

f_{s} = \mu_{s}mg

where

\mu_{s} = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

F_{net} = f_{s} - F_{c}  

0 = f_{s} - F_{c}  

\mu_{s}mg = \frac{mv^{2}}{R}  

\mu_{s} = \frac{v^{2}}{gR}  

\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544  

3 0
4 years ago
A mass of gas occupies 20 cm3 at 5 degrees Celcius and 760 mm Hg pressure. What is its volume at 30 degrees Celcius and 800 mm H
suter [353]

Answer:

20.7 cm³

Explanation:

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

Initial volume (V1) = 20 cm³

Initial temperature (T1) = 5 °C

Initial pressure (P1) = 760 mm Hg

Final temperature (T2) = 30 °C

Final pressure (P2) = 800 mm Hg

Final volume (V2) =?

Next we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

Temperature (K) = Temperature (°C) + 273

Initial temperature (T1) = 5 °C

Initial temperature (T1) = 5 °C + 273 = 278 K.

Final temperature (T2) = 30 °C

Final temperature (T2) = 30 °C + 273 = 303 K.

Finally, we shall determine the new volume of the gas as follow:

Initial volume (V1) = 20 cm³

Initial temperature (T1) = 278 K

Initial pressure (P1) = 760 mm Hg

Final temperature (T2) = 303 K

Final pressure (P2) = 800 mm Hg

Final volume (V2) =?

P1V1/T1 = P2V2/T2

760 × 20 /278 = 800 × V2/303

Cross multiply

278 × 800 × V2 = 760 × 20 × 303

Divide both side by 278 × 800

V2 = (760 × 20 × 303) / (278 × 800)

V2 = 20.7 cm³

Therefore, the new volume of the gas is 20.7 cm³

3 0
4 years ago
Is grass green ,,,,,
MatroZZZ [7]

Answer:

Yes and the sky is blue.

7 0
3 years ago
Read 2 more answers
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