All categories

2 years ago

How can stretching affect the range of motion of the neck? Hypothesis

Physics

1 answer:

kari74 [83]2 years ago

7 0

Answer:

reduce passive stiffness and increase range of movement during exercise.

Explanation:

stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.

You might be interested in

A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou

irakobra [83]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.

3 0

3 years ago

Read 2 more answers

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Yuri [45]

P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]

3 0

3 years ago

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

$v_r$ = Speed of river

$v_c$ = Speed of canoe

$v_r+v_c=2.8\ m/s$

$v_r-v_c=-1.9\ m/s$

Adding the equations we get

$2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s$

$0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s$

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

A small plane flies 37.0 km in a direction 45° north of east and then flies 28.0 km in a direction 25° north of east.

Karo-lina-s [1.5K]

Answer:

d= 64.1 km θ = 36.4º

Explanation:

a) In order to find the plane's straight-line distance from the starting point, we need to know the coordinates of the final and initial position of the plane, so we can find the total displacement, as the difference between the final and initial position.

If we choose to put our origin at the initial point of trajectory, we have that (x₀, y₀) = (0, 0)

In order to find the position of the plane after finishing the flight, we need to find its final coordinates (x₁, y₁).

In order to get x₁, we need to add the x-coordinate after flying 45º north of east, and the Δx after completing the flight in a direction 25º of east, that we can find applying trigonometry, as follows:

x₁ = 37.0 km * cos 45º + 28.0 km* cos 25º = 51.6 Km

Appying the same considerations for the y-coordinate, we have:

y₁ = 37.0 km * sin 45º + 28.0 km* sin 25º = 38.0 km

Now, as the initial position coincides with the origin, the distance in a straight line from this point to the origin, is just the hypotenuse of the triangle determined by the coordinates (x₁, y₁) and (0,0), as follows:

$d = \sqrt{x1^{2}+y1^{2}} =\sqrt{(51.6km)^{2}+(38km)^{2}} =64.1 km$

The geographic direction of the displacement vector (which coincides in magnitude with the distance we have just found), is just the angle that this distance forms with the east axis, that we can find getting the tangent of this angle as follows:

tg θ = $\frac{y1}{x1} = \frac{38km}{56.1km} =0.736$

⇒ θ = 36.4º North of East (counterclockwise from the east axis).

7 0

2 years ago