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Serhud [2]
3 years ago
9

Which of the following is an example of energy being transferred by sound waves?

Physics
2 answers:
worty [1.4K]3 years ago
8 0
First one i think nigs
Deffense [45]3 years ago
8 0

Answer:

all of them are right

Explanation:

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The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there
jenyasd209 [6]

Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

Area of crossection, A = 13 cm²

Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

e=\frac{NBA}{t}

Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

4 0
3 years ago
This mutation blank alter the protein function
mariarad [96]

Answer:

magsagot kau ng inyo wag kau mag hanap tseee hahahahhahahahahabababab

4 0
2 years ago
Calculating Net Force
VARVARA [1.3K]

Explanation:

225=m (2.20m/s2 which give m=16kg I used newtons second law to fins the required average force

6 0
3 years ago
Write four (4) strategies on how can we manage land use around South Florida. For each strategy, explain it's benefits to the en
a_sh-v [17]

Answer:

What are some ways we can improve on our land use?

- Avoid deforestation and close the forest frontier.

The first-order priority is to end deforestation by closing the "forest frontier" or intact forests, to development.

- Increase agricultural productivity.

- Restore forests and landscapes.

- Reduce food loss and waste.

- Improve diets

3 0
3 years ago
Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7
inysia [295]

Answer:

 Electric field E = kQ/r^2  

Distance between charges = 6.30 - (-4.40) = 10.70m  

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.  

Field from q1 at P = k(-9.50x^10^-6) / d^2  

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2  

These fields are in opposite directions and are equal magnitudes if the resultant field = 0  

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2  

9.50 / d^2 =8.40 / (10.70-d)^2  

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131  

d/(10.70-d) = sqrt(1.1331) = 1.063  

d = 1.063 ((10.70-d)  

= 10.63 - 1.063d  

2.063d = 10.63  

d = 5.15m  

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Explanation:

4 0
3 years ago
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