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Brrunno [24]
3 years ago
6

How long should you hold a stretch? 10-30 seconds 90 seconds 5 minutes

Physics
2 answers:
AlladinOne [14]3 years ago
6 0
The National Academy of Sports Medicine recommends holding a static stretch for about 30 seconds to achieve better flexibility. so the first one basically
galina1969 [7]3 years ago
5 0

I think you should hold a stretch for 10-30 seconds

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A man leave his townA at 8 am for town B which is 384km away and he reach at4pm what is average speed in m/s
zlopas [31]

Total distance covered = 384 Km

Total time taken to travel from A to B = 8 hours [from 8 am to 4 pm, there are 8 hours]

We know, Average speed = Total Distance Travelled/ Total Time Taken

Therefore, average speed = 384 Km/8 h = 384000m/8×60×60s =(384000/28800)m/s

= 13.3 m/s

Answer is 13.3 m/s

5 0
2 years ago
A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. T
slega [8]

Answer:

 N_s\approx41667 \hspace{3}lo ops

Explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings

In this case:

V_p=120V\\V_s=100kV=100000V\\N_p=50

Therefore, using the previous equation and the data provided, let's solve for N_s :

N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps

Hence, the number of loops in the secondary is approximately 41667.

3 0
3 years ago
Does a wave with the high-pitched have a very low frequency
xeze [42]
No it's the opposite, ths higher the pitch the greater the frequency.
3 0
3 years ago
A glider with a mass of 2 kg is moving rightward at 1 m/s. A second glider, with a mass of 3 kg, is also moving rightward, at 5
Sophie [7]

The velocity of the second glider after the collision is 4.33 m/s rightward.

<h3>Velocity of the second glider after the collision</h3>

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

  • m₁ is mass of first glider
  • m₂ is mass of second glider
  • u₁ is initial velocity of first glider
  • u₂ is initial velocity of second glider
  • v is the final velocity of the gliders

(2)(1) + (3)(5) = (2)(2) + 3v₂

17 = 4 + 3v₂

3v₂ = 17 - 4

3v₂ = 13

v₂ = 13/3

v₂ = 4.33 m/s

Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

7 0
1 year ago
Hey! I need help with this question.<br> Will give Brainlest!!<br> Thanks in advance!
Evgesh-ka [11]
The answers are In the book
8 0
3 years ago
Read 2 more answers
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