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Lerok [7]
3 years ago
12

Another word for an amp-volt is a(n) a.ohm b.watt c.joule d.coulomb

Physics
1 answer:
creativ13 [48]3 years ago
4 0
Power = (voltage) x (current)

1 watt  =  (1 volt) x (1ampere)

           =  1 "volt-amp" or 1 "amp-volt" .
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SYW A force of 175 N is needed to keep a stationary engine of weight 640 N on wooden skids from
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Answer:

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Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of
natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00  m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

r'=\sqrt{d^2+r^2}

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

r'=\sqrt{(1.00)^2+(4.00)^2}

r'=\sqrt{1.00+16.00}

r'=4.12\ m

We need to calculate the path difference

Using formula of path difference

|r'-r|=4.12-4.00

|r'-r|=0.12\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{343}{300}

\lambda=1.143\ m

We need to calculate the phase difference

Using formula of phase difference

\phi=\dfrac{2\pi\times|r'-r| }{\lambda}

\phi=\dfrac{2\pi\times0.12}{1.143}

\phi=0.659\ rad

Hence, The phase difference is 0.659 rad.

7 0
3 years ago
Which of the following is not one of the observed characteristics of our solar system?
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What is the result of a mutation during replication
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The correct answer is A. The strands are different. Hope this helps! :)

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3 years ago
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An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char
ArbitrLikvidat [17]

Answer:

Part a)

W = 1.58 \times 10^{-20} J

Part b)

v = 1.86 \times 10^5 m/s

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = 4.00 \times 10^{-12} C/m^2

now the electric field is given as

E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}

E = 0.225 N/C

Now acceleration of an electron due to this electric field is given as

a = \frac{eE}{m}

a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}

a = 3.97 \times 10^{10}

Now work done on the electron due to this electric field

W = F.d

d = 0.470 - 0.03

d = 0.44 m

So work done is given as

W = (ma)(0.44)

W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)

W = 1.58 \times 10^{-20} J

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

W = \frac{1}{2}mv^2 - 0

1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2

v = 1.86 \times 10^5 m/s

7 0
3 years ago
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