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1 year ago

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

Physics

1 answer:

kari74 [83]1 year 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.

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Hydroelectric power plants can generate huge amounts of electricity. Which of these statements best describes the impact of a hy

Zarrin [17]

The best answer is that it reduces the level of ground water

7 0

2 years ago

Read 2 more answers

Help with this question plz and show how you got it :) tysm!

spayn [35]

The hardest part of the job is to find the right formula to use, and write it down. You've already done that ! The rest is just turning the crank until an answer falls out.

You wrote. E = m g h.
Beautiful.
Now divide each side by (g h), and you'll have the formula for mass:

m = E / (g h).

You know all the numbers on the right side. Just pluggum in, do the arithmetic, and you'll have the mass.

5 0

2 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

2 years ago

What are the factors that effect the strength of a frictional force???

seropon [69]

If you write down the formula for friction, you will get an answer.

Ff = u * N Where N is a push down force that an object experiences.

u (mu) is a constant and has no units

It may not be accelerating and still experience friction. A is not correct.

Color and Density will not affect the frictional force. B is not so.

Buoyant forces are a different thing altogether. Generally friction has nothing to do with them. C is incorrect.

The last one is your answer. Technically mg should be the answer and not mass, but the second part is correct.

5 0

2 years ago

Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this l

12345 [234]

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz

5 0

2 years ago