The period of the wave is

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

A type of energy work that utilizes touch and visualization

nydimaria [60]

It would be the <span>Energy Therapies Reiki that is a common type of work wherein it makes use of touch and visualisation. In addition, this kind of therapy has its primary goal just to lessen stress and furthermore promotes relaxation. The technique is said to be making use of a "life force energy."</span>

7 0

3 years ago

Please help me with this

mezya [45]

Answer:

.067 so C

Explanation:

I asked my sister who is in 2nd grade and she said it was right so you are good! =). have a great day!

5 0

3 years ago

Calculate the minimum thickness (in nm) of an oil slick on water that appears blue when illuminated by white light perpendicular

mr_godi [17]

Answer:

The minimum thickness = 83.92 nm

Explanation:

The relation between the wavelength in a particular medium and refractive index $\lambda_n = \frac{ \lambda }{n}$

where ;

$\lambda$ = wavelength of the light in vacuum

n = refractive index of medium with respect to vacuum

For one phase change :

$2t = \frac{\lambda_n}{2}\\\\where \ \lambda_n = \frac{\lambda}{n}\\\\Then \ \\\\2t = \frac{\lambda}{2n}\\\\t = \frac{\lambda_n}{4n}$

Replacing 1.43 for n and 480 nm for λ; we have:

$t = \frac{480}{4(1.43)}$

t = 83.92 nm

Thus; the minimum thickness = 83.92 nm

4 0

3 years ago

an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

4 0

2 years ago