All categories

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.

You might be interested in

A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

Flura [38]

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

$Z = \sqrt{R^{2}+X_{L}^{2}}$

$Z = \sqrt{20^{2}+10^{2}}$

Z = 22.36 ohm

$V_{rms}=\frac{V_{0}}{\sqrt{2}}$

$V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A$

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

6 0

3 years ago

It is necessary to determine the specific heat of an unknown object. The mass of the object is 201.0 g. It is determined experim

navik [9.2K]

Mass = 0.201kg
Energy = 15J
temperature change = 10C

Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)

we can rearrange this to make specific heat capacity the subject

c = $\frac{E}{m\theta}$

c = $\frac{15}{2.01}$
c =7.46268657

6 0

3 years ago

Read 2 more answers

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

erma4kov [3.2K]

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

7 0

2 years ago

Explain how gravity and inertia interact to cause the Earth to revolve around the Sun.

eduard

Answer:

The gravity of the sun and the planets works together with the inertia to create the orbits and keep them consistent. The gravity pulls the sun and the planets together, while keeping them apart. The inertia provides the tendency to maintain speed and keep moving. The planets want to keep moving in a straight line because of the physics of inertia. However, the gravitational pull wants to change the motion to pull the planets into the core of the sun. Together, this creates a rounded orbit as a form of compromise between the two forces.

Explanation:

Hope this answer helps you....

5 0

4 years ago