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2 years ago

If the density of a diamond is 3.5 /cm^ 3 what would be the mass of a diamond whose volume is 0.5 cm ^ 3 ?

Physics

1 answer:

Elan Coil [88]2 years ago

7 0

Answer:

1.75g

Explanation:

Given parameters:

Density = 3.5g/cm³

Volume = 0.5cm³

Unknown:

Mass of the diamond = ?

Solution:

Density is the mass per unit volume of a substance

Density = $\frac{mass}{volume}$

So;

Mass = 3.5 x 0.5 = 1.75g

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What is the weight of a dog with a mass of 12 kg? Explain how you obtained your answer.

butalik [34]

Answer: 117.60N

Explanation:

Weight is a force. Therefore, we can use the force formula to find weight.

$Formula: W=m*g$

W = weight

m = mass

g = acceleration due to gravity ( $9.8m/s^2$ )

$W=(12kg)(9.8m/s^2)\\W=117.60N$

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3 years ago

Read 2 more answers

In an internal combustion engine, heat flow into a gas causes it to ......

ivann1987 [24]

In an internal combustion engine, heat flow into a gas causes it to expand.
The application of direct force to specific parts of the engine will produce expansion of the high-temperature and high-pressure gases. Which will transform the chemical energy from the fuel (such as gasoline or oi) into mechanical energy.

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3 years ago

A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the

dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

8 0

2 years ago

If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what

Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

Δm = ∑ $| \frac{dm}{dx_i} | \ \Delta x_i$

Physical quantities are precise values of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the cases, all the errors add up.

If m is the calculated quantity, x_i the measured values and Δx_i the uncertainty of each value, the total uncertainty is

Δm = ∑ $| \frac{dm}{dx_i } | \ \Delta x_i$ | dm / dx_i | Dx_i

for instance:

If the magnitude is a average of two magnitudes measured each with a different error

m = $\frac{m_1+m_2}{2}$

Δm = | $\frac{dm}{dx_1}$ | Δx₁ + | $\frac{dm}{dx_2}$ | Δx₂

$\frac{dm}{dx_1}$ = ½

$\frac{dm}{dx_2}$ = ½

Δm = $\frac{1}{2}$ Δx₁ + ½ Δx₂

Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

Learn more about propagation errors here:

brainly.com/question/17175455

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2 years ago

(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t

Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

Lf = 333,000 J/kg

Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0

3 years ago