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3 years ago

8

A sample of an unknown liquid has a volume of 24.0 mL and a mass of 6 g. What is its density? Show your work or explain how you

determined this value.

2 answers:

jek_recluse [69]3 years ago

8 0

Answer: 0.25 grams per milliliter

Explanation:

The formula to calculate the density is given by :-

$\text{Density}=\dfrac{\text{Mass}}{\text{Volume}}$

We are given that the volume of liquid is 24.0 mL.

Mass = 6g

Now, the density of liquid is given by :-

$\text{Density}=\dfrac{6}{24}\\\\\Rightarrow\ \text{Density}=0.25\ g/ml$

Hence, its density = 0.25 grams per milliliter.

adoni [48]3 years ago

3 0

I think this might be the answer let me know 0.2 g/cm^3

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Answer:

187.38 m

Explanation:

Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

Where s = distance of fall, u = initial velocity of the rock, t = time taken for the rock to fall from rest, g = acceleration due to gravity of venus.

Given: u = 0 m/s ( from rest), t = 6.5 s, g = 8.87 m/s².

substituting into equation 1

s = 0(6.5) + 1/2(8.87)(6.5)²

s = 0 + 374.7575/2

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What will happen to the speed of an object if the net force is in the direction of the motion?

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How would you change the distance between two charged particles to increase the electric force between them by a factor of 16

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The electrostatic force between two charges is inversely
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So if you want to multiply the force by, say, ' Q ',
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A solid disk has a mass of 162 kg and a radius of 1.30m. This

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Explanation:

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3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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3 years ago