What is the mass in rams of 30.00 ml of a liquid that has a density of 1.21 g/ml?

Sodium chloride and calcium carbonate combine to form sodium carbonate and calcium chloride in the reversible reaction.

lidiya [134]

Answer:

I think is Number 2

The rate at which calcium chloride is equal to the rate at which sodium chloride is produced.

Let me know if I'm wrong.

6 0

2 years ago

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A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va

natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor $C= 2.00\mu F$

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

$I = \dfrac{V}{Z}$

$Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}$

Put the value of Z into the formula

$I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}$

Put the value into the formula

$0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}$

$L=35.8\ mH$

Hence, The inductance of the inductor is 35.8 mH

4 0

3 years ago

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Concept Development

Ostrovityanka [42]

Answer:

The answer is below

Explanation:

The main difference between a liquid and a gas is that when a liquid is under pressure, its volume "won't change apparently. The reason is that the distance between the molecules of a liquid is relatively small, and the molecules of a liquid extensively withstand the compressive forces. This is similar to the distance between the molecules of a solid."

3 0

3 years ago

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

nirvana33 [79]

Answer:

4 times greater

Explanation:

Step 1: Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² = $\frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 20-m

$A_{1} = \frac{\pi d^{2}}{4}$

$A_{1} = \frac{\pi (20^{2})}{4}$

A₁ = 314.2 m²

Step 2: Calculate light-collecting area of a 10-meter Keck telescope (A₂)

$A_{2} = \frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 10-m

$A_{2} = \frac{\pi (10^{2})}{4}$

A₂ = 78.55 m²

Step 3: divide A₁ by A₂

$= \frac{314.2 m^2}{78.55 m^2}$

= 4

Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0

3 years ago

1) What is the weight of a 356kg object?

stiks02 [169]

1) 3489 N

2) The microbumps over the surfaces

3) See explanation

Explanation:

1)

The gravitational force acting on an object (also known as weight of the object) is the attractive force with which the Earth pulls the object towards the ground.

It can be calculated using the formula:

$W=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 356 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the weight of the object is:

$W=(356)(9.8)=3489 N$

2)

The force of friction is a force that acts whenever there is an object sliding over a surface. It is a resistive force, therefore its direction is always opposite to the direction of motion of the object.

On a microscopic scale, the reason of this force of friction is the presence of "microbumps" over the surface of the sliding object (and over the surface of the floor). Because of the roughness of the two surfaces, the molecules of the two surfaces "interact" and cause a resistive force that opposes the relative motion of the object over the surface.

Macroscopically, for an object over a flat surface, the magnitude of the force of friction is

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

3)

There are two types of force of friction:

- Force of static friction: this occurs when the object is still at rest. We experience this force when we try to put in motion an object on the ground, by applying a push: we notice that if the force we apply is not strong enough, the object will remain at rest. This is because there is a static force of friction acting on the object, that equals the force of push that we are applying, but in the opposite direction. The magnitude of the force of friction can vary up to a maximum value, given by

$F_{max}=\mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the object

g is the acceleration due to gravity

- Force of kinetic friction: this is the force of friction that occurs when the object is already in motion. The kinetic force of friction opposes the motion of the object, and its magnitude is

$F_k = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

Generally, for most surfaces, $\mu_k < \mu_s$ , therefore the force of kinetic friction is generally less than the force of static friction.

4 0

3 years ago