Use the information from the graph to answer the

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

3 years ago

A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit

Alexus [3.1K]

Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

$10^5 Q +300Q'+0.2Q''=0$

With the initial conditions as,

$Q(0) = 10^{-6}$

$Q'(0)= 0$

Where Q(t) is the charge.

The general solution of a linear equation is given as:

$y(x) = c_1e^{-ax}+c_2e^{-bx}$

Applying this definiton in our differential equation we have that

$Q(t) = C_1e^{at}+C_2e^{bt}$

To find b and a we use the first equation and find the roots:

$r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}$

$r_{a,b} = {-1000,-500}$

Then we have

$Q(t) = C_1e^{-1000t}+C_2e^{-500t}$

To find the values of the Constant we apply the initial conditions, then

$Q(0)= 10^{-6} = C_1+C_2$

And for the derivate:

$Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}$

$0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}$

$0 = -1000C_1-500C_2$

We have a system of 2x2:

$(1) 10^{-6} = C_1+C_2$

$(2) 0 = -1000C_1-500C_2$

Solving we have:

$C_1 = -10^{-6}$

$C_2 = 2*10^{-6}$

The we can replace at the equation and we have that the Charge at any moment is given by,

$Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}$

If we obtain the derivate we find also the Current, then

$I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}$

7 0

3 years ago

When large amplitude sound vibrates the ear significantly, a _______ results.

kondaur [170]

D loud sound

Hope this helped:)

4 0

3 years ago

Read 2 more answers

Is god real??? i wanna know

NemiM [27]

In my opinion, yes the bible tell us that "For God so loved the world that he gave his one and only Son, that whoever believes in him shall not perish but have eternal life"
So my answer is yes

6 0

3 years ago

Read 2 more answers

If a car moves with constant velocity, does it also move with a constant speed?

GarryVolchara [31]

Answer:

Yes. The fact that an object moves at constant velocity implies that its speed is also constant. Note that the converse statement isn't necessarily true.

Explanation:

Velocity is a vector. For two vectors to be equal to each other,

their magnitudes (sizes) need be the same, and
they need to point in the same direction.

In motions, the magnitude of an object's velocity is the same as its speed.

If the car moves with a constant velocity, that means that

the magnitude of its velocity, the speed of the car, is constant;
also, the direction of the car's motion is also constant.

In other words,

$\text{Velocity } \vec{v}\text{ is constant} \implies \text{Speed }v\text{ is constant}$ .

Note that the arrow here points only from the velocity side to the speed side. It doesn't point backward because knowing that the speed of an object is constant won't be sufficient to prove that the velocity of the object is also constant. For example, for an object in a uniform circular motion, the speed is constant but the direction keeps changing. Hence the velocity isn't constant.

3 0

3 years ago