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monitta
3 years ago
9

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Physics
1 answer:
Korvikt [17]3 years ago
6 0

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h  

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3  

V = 6.307 x 10^-5 m^3  

By density = m/V  

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

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Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by v_{max}=A\omega where A is amplitude and \omega is angular velocity

Angular velocity is given by \omega=\sqrt{\frac{k}{m}}  where k is spring constant and m is mass

So v_{max}=A\sqrt{\frac{k}{m}}

A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m

3 0
3 years ago
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We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans
Ksivusya [100]

Answer:

Range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

Explanation:

We have given range of frequency is 400-560 Hz

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We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by v=\lambda f, here \lambda is wavelength and f is frequency

So for 400 Hz frequency wavelength will be \lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m

And wavelength for frequency 560 Hz \lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m

So range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

8 0
3 years ago
A gas always spreads out to fill all available space.<br> True or False <br> Science Hw
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Answer:

true

Explanation:

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2. Which of the following is accurate when discussing specific neat?
MrRissso [65]

Answer:

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3 years ago
A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from
o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}

There is NO initial vertical velocity, so:

\large\boxed{\Delta d= \frac{1}{2}at^2}}

Rearrange to solve for time:

2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

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Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

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3 years ago
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