A molecule of water (H2O) is made from _____ combining two hydrogen atoms and one oxygen atom.

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

True or false the potential energy of a freely object increases as it begins to fall

Eva8 [605]

False, as an object falls its potential energy turns into kinetic energy thus decreasing the potential energy.

3 0

3 years ago

The 6 strings on a guitar all have about the same length and are stretched with about the same tension.The highest string vibrat

ahrayia [7]

Answer:

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

Frequency can be further expanded to;

f = v/2L = (1/2L)√(T/u) ......1

Where;

m= mass,

u = linear density of string,

T = tension

p = density of string material

A = cross sectional area of string

d = string diameter

u = m/L .......2

m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)

f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))

f = (1/2L)√(T/((p(πd^2)L/4)/L))

f = (1/2L)√(4T/pπd^2)

f = (1/L)(1/d)√(4T/pπ)

Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

f ~ 1/d

So, if

4f(low) = f(high)

Then,

d(low) = 4d(high)

6 0

3 years ago

What will occur when the trough of Wave A overlaps the trough of Wave B?

Pavel [41]

C because the wave length was longer

6 0

3 years ago

Read 2 more answers

Find the average energy E for (a) An n-state system, in which a given state can have energy 0,,2,...,n . (b) A harmonic oscillat

kicyunya [14]

Answer:

a♦1 E_average = n E₀ / 2 , b) E_average= infinity

Explanation:

The energy values form an arithmetic series, whose sum is

S = n (a₁ + aₙ) / 2 = n (2a₁ + (n-1) r)/ 2

Where n is the number of terms, a₁ is the first term, aₙ the last term and r is the difference between two consecutive numbers in the series

r = 2E₀ - 0 = 2E₀

Therefore the sum is

S = n (0 + n E₀) / 2

S = n² E₀ / 2

The average value is

E_average = S / n

E_average = n E₀ / 2

b) the case of harmonic oscillation

We have two possibilities.

- if we take a finite number and terms gives the same previous value

- If we take an infinite number of fears the series gives infinity and the average is also infinite

E_average= infinity

4 0

3 years ago