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3 years ago

You hang a 350 gram mass on a spring and its displaced 0.5 cm. Its spring constant is

Physics

1 answer:

Keith_Richards [23]3 years ago

4 0

Answer:

Explanation:

kx = mg

k = mg/x

k = 0.350(9.81)/0.005

k = 686.7

k = 687 N/m

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Janet jumps horizontally off a high diving platform with a velocity of 2.63 m/s and lands in the water 1.9 s later. How high is

soldier1979 [14.2K]

Answer:

The height of the platform is 22.686 m.

Explanation:

Given;

initial velocity of Janet, u = 2.63 m/s

time of her motion, t = 1.9 s

The height of the platform is given by;

h = ut + ¹/₂gt²

substitute the given values and solve for height "h"

h = (2.63 x 1.9) + ¹/₂(9.8)(1.9)²

h = 4.997 + 17.689

h = 22.686 m

Therefore, the height of the platform is 22.686 m.

5 0

3 years ago

What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.

jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = $1 \times 10-12 W/m^2$

Now

Intensity level ( or Loudness)is

$L = log10 \frac{I}{10}$

$L = log10 \frac{10}{1\times 10^{-12}}$

$L = log10 \times 1013$

$= 13 \times 1 ( log10(10) = 1)$

Therefore

L = 13 bel

And as we know that

1 bel = 10 decibels

So,

The Sound intensity level is

L = 130 decibels

5 0

3 years ago

How do the meetings of the words exercise and fitness differ?

ICE Princess25 [194]

Exercise is the activity and fitness is a lifestyle and done with time

7 0

3 years ago

Why does the light in space travel like a wave

Thepotemich [5.8K]

Answer:

basic answer, radiation.

7 0

3 years ago

Read 2 more answers

Aluminum rivets used in airplane construction are made slightly larger than the rivet holes and cooled by "dry ice" (solid CO2)

poizon [28]

Answer:

$\rm 4.554\ m m.$

Explanation:

<u>Given:</u>

Initial temperature, $\rm T_i=23.0\ ^\circ C=23 + 273.15 = 296.15\ K.$
Final temperature, $\rm T_f = -78.0\ ^\circ C = -78+273.15=195.15\ K.$
Diameter of the hole, $\rm d = 4.500\ m = 4.500\times 10^{-3}\ m.$
Expansion coefficient of the Aluminum, $\rm \alpha = 2.4\times 10^{-5}\ K^{-1}.$

Let the diameter of the rivet at $\rm T_i=23.0\ ^\circ C$ be $\rm d_o$ , such that,

$\rm d_o=d+\Delta d$

$\rm \Delta d$ is the elongation in the diameter of the rivet.

We know, The change in the diameter of the rivet is related with the change in temperature as:

$\rm \dfrac{\Delta d}{d_o}=\alpha \Delta T.$

where, $\rm \Delta T$ is the change in temperature = $\rm T_f-T_i=-195.15-296.15=-491.30\ K.$

Also, $\rm \Delta d=d_o-d.$

Using these values,

$\rm \dfrac{d_o-d}{d_o}=\alpha \Delta T\\1-\dfrac{d}{d_o}=\alpha \Delta T\\\dfrac{d}{d_o}=1+\alpha \Delta T\\\Rightarrow d_o = \dfrac{d}{1+\alpha \Delta T}\\=\dfrac{4.500\times 10^{-3}}{1+2.4\times10^{-5}\times (-491.30)}\\=4.554\times 10^{-3}\ m.\\=4.554\ m m.$

It is the required diameter of the rivet at $-78.0\ ^\circ C$ .

4 0

3 years ago