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2 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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The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th

Bezzdna [24]

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

7 0

2 years ago

What is the average speed of a car that moved 40 km in 3 hours?

Naya [18.7K]

<u>A</u> would be the answer, since it would take about 3 hours to cover most of it, 10km/h would be the average speed.

really hope this helps.

3 0

3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

2 years ago

Atoms are happy (they will not readily react with other elements) when they have a full outside ring of

Len [333]

Answer: TRUE

Explanation:

Atoms are happy when they will not react with other elements while having a full outside ring of electrons because this makes them to be noble.

A stable atom possesses full outside ring of electrons while unstable one does not. So, they are happy also because of stability.

7 0

2 years ago

A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference

Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

$q=CV$

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

$C'=kC$

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

$q=C'V'$

So we can combine the two equations:

$CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}$

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

$V'=\frac{71.0}{4.50}=15.8 V$

6 0

2 years ago