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4 years ago

Is the following equation balanced? SO3 + 2H2O H2SO4 yes no

Physics

2 answers:

sergeinik [125]4 years ago

8 0

Answer:

Explanation:

it is not balanced because there are four oxygen atoms in H2SO4 while there 5 oxygen atoms in the reactants side. There are also more hydrogen atoms on the reactants side. To make it balanced remove the 2 in front of H2O. :)

Gnom [1K]4 years ago

5 0

No it is not balanced because there are four oxygen atoms in H2SO4 while there 5 oxygen atoms in the reactants side. There are also more hydrogen atoms on the reactants side. To make it balanced remove the 2 in front of H2O. :)

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On a distance-time graph, time is shown on the y-axis.<br> A) True<br> B) False

Artist 52 [7]

Answer:

false : In distance time graph,time is shown on the x -axis

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3 years ago

A proposed space station includes living quarters in a circular ring 50.0 m in diameter. At what angular speed should the ring r

mel-nik [20]

Given that the space station is free of gravitational force, it is required that it spins an certain speed to acquire centripetal acceleration.

In this case, you want that the centripetal acceleration, Ac, equals g (gravitational acceleration on the earth), becasue this will cause a centripetal force equal to the weight on earth.

The formula for centripetal acceleration is Ac = [angular velocity]^2*R

where R = [1/2]50.0m = 25.0 m

Ac = 9.81 m/s^2

=> [angular velocity]^2 = Ac/R = 9.81m/s^2v/ 25.0m = 0.3924 (rad/s)^2

[angular velocity] = √(0.3924) rad/s = 0.63 rad/s

Answer: 0.63 rad/s

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3 years ago

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.

oksano4ka [1.4K]

Answer:

$i = 2.84 \times 10^{-3} A$

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

$J = \frac{di}{dA}$

so the current through the wire is given as

$i = \int J dA$

now we have

$i = \int_{0.921R}^R J dA$

here we have

$J = (3.25 \times 10^8)r^2$

now plug in the values in above equation

$i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr$

now we have

$i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr$

$i = (2.04 \times 10^9) \frac{r^4}{4}$

now plug in both limits as mentioned

$i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})$

$i = (2.04\times 10^9)(0.07 R^4)$

here R = 2.11 mm

$i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)$

$i = 2.84 \times 10^{-3} A$

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3 years ago

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liraira [26]

Brass, Salt and water are examples of a Homogeneous Mixture.

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I am Lyosha [343]

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3 years ago