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3 years ago

Gold has a density of 19.3 grams per cubic centimeter. what is the density of gold in metric tons per cubic meter?

1 answer:

6 0

In order to determine this, we will first need some conversions. We will need to convert metric tons and grams into one another and also cubic centimeters to cubic meters into one another.

1 metric ton = 1000 kg

1 kg = 1000 grams

1 metric ton = 10⁶ grams

So 10⁶ grams / metric ton

1 meter = 100 cm
1 m³ = (100)³ cm³
1 m³ = 10⁶ cm³

So 10⁶ cm⁶ / m³

Now, we manipulate the given value:

(19.3 grams / cm³) * (1 metric ton / 10⁶ grams) * (10⁶ cm³ / m³)
= 19.3 metric tons / m³

The density of gold is 19.3 metric tons meter meter cubed.

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A ball is spun around in circular motion such that it completes 50 rotations in 25 s. What is the frequency of its rotation? 2.

Elis [28]

Answer:

1. $f = 2 Hz$

2. $f = 0.011 Hz$

3. $f = 0.067 Hz$

4. $t = 4 s$

Explanation:

1. The frequency of rotation is given by:

$f = \frac{\omega}{2\pi}$

Where:

ω: is the angular speed = 50 rotations (revolutions) in 25 s.

We need to convert the units of ω.

$\omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s$

Now, the frequency is:

$f = \frac{4\pi rad/s}{2\pi} = 2 Hz$

2. The frequency is:

We know:

5 laps = 5 revolutions

t: time = 450 s

$f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz$

3. The frequency of the pendulum is:

$f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz$

4. We have:

θ: number of revolutions = 48 rev

f = 12 Hz

t =?

The time can be calculated as follows:

$f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t}$

$t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s$

I hope it helps you!

4 0

3 years ago

What statement most accurately classifies this cell?

iogann1982 [59]

Answer:

D It is an animal cell.

Explanation:

6 0

4 years ago

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

trapecia [35]

Answer:N=13.53 rpm

Explanation:

Given

radius $r=7.80 m$

Centripetal acceleration $a_c=1.6 g$

and centripetal acceleration $a_c=\omega ^2r$

where $\omega =angular\ velocity$

$1.6\times 9.8=\omega ^2\times 7.8$

$\omega ^2=2.0102$

$\omega =1.417 rad/s$

and $\omega =\frac{2\pi \cdot N}{60}$

$1.417\times 60=2\pi \cdot N$

$N=13.53 rpm$

8 0

3 years ago

A 0.22 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.13 m .

andre [41]

Answer:

2.3687599 m/s

0.91106 m/s

0.617213012 J

Explanation:

f = Frequency = $2.9\ Hz$

A = Amplitude = 0.13 m

k = Spring constant

m = Mass of object = 0.22 kg

Angular velocity is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 2.9\\\Rightarrow \omega=18.22123\ rad/s$

Velocity is given by

$V=A\omega\\\Rightarrow V=0.13\times 18.22123\\\Rightarrow V=2.3687599\ m/s$

Speed when it passes the equilibrium point is 2.3687599 m/s

Frequency is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow k=m4\pi^2f^2 \\\Rightarrow k=0.22\times 4\pi^2\times 2.9^2\\\Rightarrow k=73.04296\ N/m$

x = Displacement = 0.12 m

In this system the energies are conserved

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{73.04296(0.13^2-0.12^2)}{0.22}}\\\Rightarrow v=0.91106\ m/s$

The speed when it is 0.12 m from equilibrium is 0.91106 m/s

The energy in the system is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}\times 73.04296\times 0.13^2\\\Rightarrow E=0.617213012\ J$

The total energy of the system is 0.617213012 J

4 0

4 years ago

Convert 10095 mm to pm

ludmilkaskok [199]

Answer:

1.01 × 1013 picometres

Explanation:

multiply the length value by 1e+9

3 0

3 years ago