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creativ13 [48]
3 years ago
15

Answer plzzHow do we apply Interval Training to our lower body for a workout?

Physics
1 answer:
weqwewe [10]3 years ago
5 0
Interval training is simply alternating short bursts (about 30 seconds) of intense activity with longer intervals (about 1 to 2 minutes) of less intense activity. For instance, if your exercise is walking and you're in good shape, you might add short bursts of jogging into your regular brisk walks.
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If some raps 3words per second then how many words would he say in 9,999,999,990 seconds ​
wolverine [178]

Answer:

29,999,999,970 words

Explanation:

9,999,999,990x3

3 0
3 years ago
Read 2 more answers
Which machine can create images of atoms?
Svetradugi [14.3K]

Answer: Scanning Tunneling Microscope

Explanation:

The answer is, of course, the greatest Star Trek fan art imaginable: images literally built out of individual atoms. The images are the work of IBM scientists who created the unique artwork with a two-ton machine called the Scanning Tunneling Microscope that moves single atoms across a tiny piece of copper.

6 0
2 years ago
A person who weighs 509,45 N empties her lungs as much as
Masja [62]

Answer:

The weight of the girl = 1045.86 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.

From Archimedes principle,

R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.

⇒ D₁/D₂ = W/U............................... Equation 1.

Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.

Making D₁ the subject of the equation,

D₁ = D₂(W/U)................................... Equation 2

<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>

<em>Substituting these values into equation 2</em>

D₁ = 1000(509.45/487.11)

D₁ = 1045.86 kg/m³

Thus the weight of the girl = 1045.86 kg/m³

<em></em>

7 0
3 years ago
Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
Read 2 more answers
What happens when light enters a pair of glasses
alexdok [17]

Answer:

It refracts when it hits the glass.

6 0
3 years ago
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