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3 years ago

A hydrated crystal that has a water vapor pressure greater than the water vapor pressure of air is called _____.

1 answer:

6 0

The answer is Efflorescent hydrate. <span>They are chemical substances that, when exposed to the atmosphere, release water. A typical example of this chemical reaction is cement, when mixed with water, the mixture hardens due to hydration, then the cement starts to lose water.</span>

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What is the equivalent resistance of the

BigorU [14]

Answer:

Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

Since $R_{1}$ and $R_{2}$ are connected in parallel, the voltage across the two resistors would both be $V$ . Thus, the current going through the two resistors would be $(V / R_{1})$ and $(V / R_{2})$ , respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .

7 0

1 year ago

In order to transmit information via radio waves, the waves need to be changed somehow. For car radios this can happen in two wa

laila [671]

Answer: amplitude

Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres

Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.

7 0

3 years ago

A 600 kg car is at test and then accelerated to 5m/s , what is its original kinetic energy

ycow [4]

Answer:

0 J

Explanation:

Kinetic energy is defined as:

KE = 1/2 m v²

where m is mass and v is velocity.

The car starts at rest, so it has zero velocity. Therefore, its initial kinetic energy is 0 J.

3 0

2 years ago

A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

$a=0$

This means that the net force is also zero:

$\sum F=0$

The net force on the block is given by the applied force, forward, and the frictional force, backward:

$\sum F = F_a-F_f=0$

where

$F_a=39.4 N$ is the applied force

$F_f$ is the frictional force

Therefore, solving for $F_f$ ,

$F_f=F_a=39.4 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0

2 years ago

What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i

Katarina [22]

R = 0.407Ω.

The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0

2 years ago