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4 years ago

If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?

Physics

1 answer:

Alex Ar [27]4 years ago

6 0

Answer:

They have same density

Explanation:

The density of an object is defined as

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

Let's call $m_c$ and $V_c$ the mass and the volume of ball C, respectively. Therefore, the density of ball C is:

$d_c = \frac{m_c}{V_c}$

We know that the volume of ball C is 3 times the volume of ball D, so

$V_c = 3 V_d \rightarrow V_d = \frac{V_c}{3}$

And we also know that ball D has 1/3 the mass of ball C:

$m_d = \frac{m_c}{3}$

So, the density of ball D is:

$v_d = \frac{m_d}{V_d}=\frac{m_c/3}{V_d/3}=\frac{m_c}{V_c}=d_c$

Therefore, the two balls have same density.

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Describe how an electromagnet can be switched on and off.

Phoenix [80]

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

8 0

3 years ago

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication

koban [17]

Answer:

r = 4.24x10⁴ km.

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>

From equation (1), r₁ is:

$r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}$

$r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}$

$r_{1} = 4.24 \cdot 10^{4} km$

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0

4 years ago

You have 13.0 gallons of gasoline. How many liters is this? (1 gallon equals<br> about 3.79 L.)

Harlamova29_29 [7]

Answer:

I believe the answer is C

Explanation:

8 0

4 years ago

Read 2 more answers

Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.

anzhelika [568]

Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
$E^0=E_{cat}^0-E_{an}^0$
where $E_{cat}^0$ is the standard potential at the cathode, while $E_{an}^0$ is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: $E_{Cu}^0=+0.34 V$ , while at the anode we have zinc: $E_{Zn}^0=-0.76 V$ . Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
$E^0=+0.34 V-(-0.76 V)=+1.1 V$

2) To calculate $E^0$ at any temperature T, we should use Nerst equation:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}$
where
$R=8.31 J/(K mol)$
$T=473.15 K$ is the temperature in our problem
$z=2$ is the number of electrons transferred in the cell's reaction
$F=9.65\cdot 10^4 C/mol$ is the Faraday's constant
$[Zn]$ and $[Cu]$ are the molar concentrations of zinc and in copper, and in our problem we have $[Zn]=10[Cu]$ .
Using all these data inside the equation, and using $E^0=+1.1 V$ , in the end we find:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

8 0

3 years ago

The density of iron is 7.87 g/cm3. If an iron cannonball has a volume of 105 cm3

Reptile [31]

the answer is 0.284 lb/in3

6 0

3 years ago