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3 years ago

Round 0.010229 to four sig figs

Physics

1 answer:

inysia [295]3 years ago

7 0

It would be 0.01023.

0.010229 has 5 sig. fig.'s (leading zeros do not count)

So rounding to 4 sig. fig.'s is just a matter of rounding off the last digit, which leaves you with 0.01023.

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A speaker at the front of a room and an identical speaker at the rear of the room are being driven at 456 Hz by the same sound s

vampirchik [111]

Answer:

2.97795 Hz

Explanation:

v = Speed of sound in air = 343 m/s

$v_r$ = Relative speed between the speakers and the student = 1.12 m/s

$f'$ = Actual frequency of sound = 456 Hz

Frequency of sound heard as the student moves away from one speaker

$f_1=f'\dfrac{v-v_r}{v}\\\Rightarrow f_1=456\dfrac{343-1.12}{343}\\\Rightarrow f_1=454.51102\ Hz$

Frequency of sound heard as the student moves closer to the other speaker

$f_2=f'\dfrac{v+v_r}{v}\\\Rightarrow f_2=456\dfrac{343+1.12}{343}\\\Rightarrow f_2=457.48897\ Hz$

The difference in the frequencies is

$f_2-f_1=457.48897-454.51102=2.97795\ Hz$

The student hears 2.97795 Hz

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3 years ago

Westinghouse and Edison fought what was known as the war of the currents. Eventually, Westinghouse triumped using Alternating Cu

DerKrebs [107]

The answer is B! :) Hope it helped

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3 years ago

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A book weighing 1.2 kg stands on the table. We place our hand on the book with a force of 15.3 N. The reaction force of the tabl

babunello [35]

Answer:

27 N

Explanation:

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2 years ago

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at

aliina [53]

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =\frac{k*q}{d}$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = \frac{2*q*k}{d} = \frac{2*8.99e9N*m2/C2*4e-6C}{8m} =\\ \\ V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = \frac{8.99e9N*m2/C2*(4e-6C)}{8m} + (\frac{8.99e9N*m2/C2*(-4e-6C)}{8m}) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

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3 years ago

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skad [1K]

Answer: I'm not sure, but I think it would be a <em>total lunar eclipse </em>

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3 years ago