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3 years ago

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 a sloping surface, like a ramp, that reduces the amount of force required to do work.

puteri [66]

If this is about simple machines the answer is: Inclined plane

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3 years ago

You connected the 5 Ω, 10 Ω, 15 Ω resistors in series with a 90 V battery. What is the current?

kykrilka [37]

Answer:

Explanation:

Rtoal=R1+R2+R3=5+10+15=30

I=V/R 90/30

I=3

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2 years ago

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

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3 years ago

The volume of an object is the amount of space it takes up

soldier1979 [14.2K]

Yes it is. Uh huh, uh huh, shore enuff. Mmm hmm. Yeah yeah yeah. Yah Mon ! Indubitably.

6 0

3 years ago

Read 2 more answers

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

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3 years ago

Read 2 more answers