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Dmitrij [34]
3 years ago
8

Balance the following equation Fe203+HCL fecl3+H20

Physics
1 answer:
Arturiano [62]3 years ago
3 0
Balanced equation:
6 HCl + Fe2O3 = 2 FeCl3 + 3 H2O
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What is meaning of phase​
algol [13]

Answer:

phares

phares \\  \\ means \: a \: noun \:  \\  \\ i \: thnk \: so \:  \:

5 0
2 years ago
Read 2 more answers
For the circuit shown in (Figure 1), find the potential difference between points a and b. Each resistor has
Lynna [10]

The potential difference between points a and b is zero.

<h3>Total emf of the series circuit</h3>

The total emf in the circuit is the sum of all the emf in the circuit.

emf(total) = 1.5 + 1.5 = 3.0 V

<h3>Potential difference</h3>

The potential difference between two points, a and b is calculated as follows;

V(ab) = Va - Vb

V(ab) = 1.5 - 1.5

V(ab) = 0

Thus, the potential difference between points a and b is zero.

Learn more about potential difference here: brainly.com/question/3406867

3 0
2 years ago
What is the moving kinetic energy of a 1500kg car moving at 25m/s?
EleoNora [17]

Answer:

<h2>468,750 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1500 \times  {25}^{2}  \\  = 750 \times 625

We have the final answer as

<h3>468,750 J</h3>

Hope this helps you

4 0
3 years ago
What is most likely the relationship between Mendel’s rules of genetics and Darwin’s idea of biological evolution?
xenn [34]

Answer:

Darwin's theory of natural selection lacked an adequate account of ... Darwinian principles now play a greater role in biology than ever before, .... Sadly, even if Mendel had lived to see the rediscovery of his work, he probably .... evolutionary forces are acting, a genetic equivalent to Galileo's law of inertia.

Explanation:

4 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
3 years ago
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