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3 years ago

Which activity is the best example of cardiovascular and strength training exercises work together

Physics

2 answers:

Nataly [62]3 years ago

4 0

it got to be jumping jacks that all i got

Natasha2012 [34]3 years ago

3 0

Answer:

<h2>Construction tasks.</h2>

Explanation:

A common activities where cardiovascular and strength training are together is during Construction tasks. In this context, aerobic and anaerobic exercises are present.

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What is one way that early scientific practice differed from modern scientific practice?

mote1985 [20]

Early hypotheses were not based on observations.
Early hypotheses were not tested by experimentation.
Early hypotheses were formed from scientific questions.
Early hypotheses were influenced by creative thinking

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3 years ago

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A diver jumps from a high platform and stays in the air for 1 second. How high was the platform?

ivolga24 [154]

Answer: Depends

Explanation:

Depends on how much the diver weighs.

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1 year ago

Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the

shutvik [7]

Period = 6 seconds and $frequency = 0.167Hz$ .

<u>Explanation:</u>

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of $\frac{1}{t}$ . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency = $\frac{1}{t}$

⇒ $frequency = \frac{1}{time}$

⇒ $frequency = \frac{1}{6}$

⇒ $frequency = 0.167Hz$

Therefore , Period = 6 seconds and $frequency = 0.167Hz$ .

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2 years ago

It takes 6400 years for one gram of radium to decay away to only 1/16 (one-sixteenth) of a gram. The half-life of radium is

Vsevolod [243]

1/16........................................

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2 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

Cavity and benzene should be extended in equal quantities.

$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$

$\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\$

5 0

2 years ago