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JulsSmile [24]
3 years ago
13

List 3 differences between aerobic and anaerobic exercise.

Physics
1 answer:
brilliants [131]3 years ago
6 0
Aerobic means “with air.” So things like yoga, stretching, or soft walking would be aerobic. Anaerobic means “without air.” Running up some stairs, sprinting, or lifting weights causes you to loose air quickly, making them anaerobic.

Hope that helps!
You might be interested in
How do I do this physics problem about potential energy and kinetic energy?
larisa86 [58]

Ok i apologise for the messy working but I'll try and explain my attempt at logic

Also note i ignore any air resistance for this.

First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.

Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.

I also noted the "initial" and "final" height of the swing with hi and hf respectively.

So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.

So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).

Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.

The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).

I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx

4 0
3 years ago
Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how far south does the arrow
DanielleElmas [232]

Answer:c

Explanation:its the answer because its the answer

4 0
3 years ago
What is the instantaneous velocity v of the particle at t=10.0s?
algol [13]
The instantenous velocity is just the slope of the graph at a certain instant. Since the graph is a straight line, its instantenous velocity is uniform through out. v = dx / dt = (40 - 10) / (50 - 0) = 0.6 m/s.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
6 0
3 years ago
Read 2 more answers
The space shuttle is accelerated off its launch pad to a velocity of 525 m/s in 18.0 seconds.
Eva8 [605]

Answer: 29.17m/s^2

Explanation:

Given the following :

Velocity = 525 m/s

Time = 18 seconds

Acceleration = change in Velocity with time

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity and t = time

Plugging our values

525 = 0 + a × 18

525 = 18(a)

a = 525 / 18

a = 29.166666

a = 29.17 m/s^2

8 0
3 years ago
A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th
Alex73 [517]

Answer:

V_f = 13.8863 \angle 60.89\°

Explanation:

Our values are,

m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

m_1v_1=(m_1+m_2)v_fcos\theta

For the particular case on the Y axis, we do the same with the speed of object 1.

m_2v_2=(m_1+m_2)v_fsin\theta

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°

Replacing in any of the two functions, given above, we will find the final speed after the collision,

(950)(16)=(950+1300)V_fcos(60.89)

V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}

V_f = 13.8863 \angle 60.89\°

8 0
3 years ago
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