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3 years ago

What is aerobic activities

Physics

2 answers:

Gemiola [76]3 years ago

8 0

Explanation:

Aerobic exercise is physical exercise of low to high intensity that depends primarily on the aerobic energy-generating process. "Aerobic" is defined as "relating to, involving, or requiring free oxygen", and refers to the use of oxygen to adequately meet energy demands during exercise via aerobic metabolism

luda_lava [24]3 years ago

5 0

Answer:

Aerobic exercise is any type of cardiovascular conditioning. It can include activities like brisk walking, swimming, running, or cycling. You probably know it as “cardio.”

By definition, aerobic exercise means “with oxygen.” Your breathing and heart rate will increase during aerobic activities. Aerobic exercise helps keep your heart, lungs, and circulatory system healthy.

Aerobic exercise differs from anaerobic exercise. Anaerobic exercises, such as weightlifting or sprinting, involve quick bursts of energy. They’re performed at maximum effort for a short time. This is unlike aerobic exercises. You perform aerobic exercises for a sustained period of time.

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3 years ago

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3 years ago

a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the

mihalych1998 [28]

Answer:

Part(a): The average acceleration is $1.83~m~s^{-2}$ during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if ' $v_{i}$ ', ' $v_{f}$ ' and ' $a_{av}$ ' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

$a_{av} = \dfrac{v_{f} - v_{i}}{t}$

where ' $t$ ' is the time taken by the particle to achieve the velocity ' $v_{f}$ ' starting from initial velocity ' $v_{i}$ '

Given in the problem, $v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s$ .

So the average acceleration( $a_{av}$ ) during the first 6 s will be

$a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}$

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.

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3 years ago