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3 years ago

Is stretching before working out important?

Physics

2 answers:

Damm [24]3 years ago

8 0

Answer:

Yes, you can twist or break parts of your body if you do not stretch before working out. It is always important to stretch even if you are doing activities such as gymnastics and karate.

Korolek [52]3 years ago

4 0

Answer:

yes

Explanation:

This is because stretching helps the muscles get ready before actually working out, so you don't strain or twist something.

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51.Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveli

Komok [63]

Complete question is;

Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)

(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?

Answer:

A) P_out = 24 W

B) t = 1470 s

C) Q = 1140.72 KJ

Explanation:

We are given;

Input Power; P_in = 800 W

Efficiency; η = 3% = 0.03

A) Formula for efficiency is;

η = P_out/P_in

Making P_out the subject, we have;

P_out = η•P_in

P_out = 0.03 × 800

P_out = 24 W

B) We know that;

Power = work done/time taken

Thus;

P_out = mgh/t

We are given;

m = 3000 kg

h = 1.20 m

Thus, time is;

t = (3000 × 9.8 × 1.2)/24

t = 1470 s

C) amount of heat wasted is calculated from;

Q = (P_in - P_out)t

Q = (800 - 24) × 1470

Q = 1,140,720 J

Q = 1140.72 KJ

3 0

2 years ago

Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de

-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

$s(t)^2={h^2+x^2}$ ...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

$2s(t)s'(t)=2x(t)x'(t)$

$x'(t)=\dfrac{s(t)s'(t)}{x(t)}$

When x = 40, $s(t)=\sqrt{40^2+4^2}=40.19\ m$

$x'(t)=\dfrac{-240s(t)}{x(t)}$

$x'(t)=\dfrac{-240\times 40.19}{40}$

$x'(t)=-241.14\ m/s$

So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.

8 0

3 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$