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s2008m [1.1K]
3 years ago
12

Fe + Br2 --> FeBr3

Physics
1 answer:
eimsori [14]3 years ago
6 0

Explanation:

  • 1
  • 2
  • 1
  • 1
  • 3
  • 4
  • No
  • 2Fe+3Br2-->2FeBr3
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Based on our study of electromagnets, what would be the best thing to do to a generator in order to produce more electricity?
Keith_Richards [23]

Answer:

A

Explanation:

just took a test on ed sorry if wrong I got it right

4 0
3 years ago
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The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or
Stolb23 [73]

Answer:

I = (1.80 × 10⁻¹⁰) A

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = (1.0 × 10⁻¹⁵) T

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 3.6 cm = 0.036 m

(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)

4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036

I = (1.80 × 10⁻¹⁰) A

Hope this Helps!!!

3 0
3 years ago
(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104
Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0
3 years ago
A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.
DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy Q_{out}=1000\ Btu

Work done W_{cycle}=300\ Btu

We need to calculate the coefficient of performance

Using formula of  the coefficient of performance

COP=\dftrac{Q_{in}}{W_{cycle}}

We need to calculate the Q_{in}

W_{cycle}=Q_{out}-Q_{in}

Put the value into the formula

300=1000-Q_{in}

Q_{in}=300-1000

Q_{in}=700\ Btu

Now put the value of Q_{in} into the formula of COP

COP=\dfrac{700}{300}

COP=\dfrac{7}{3}=2.33

Hence, The coefficient of performance for the cycle is 2.33.

5 0
3 years ago
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