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2 years ago

12

I stand 0.72 meters in front of a mirror and look at the person on the other side who looks like me. That person is actually you

nger than me by a small amount. How much?

1 answer:

Sergeu [11.5K]2 years ago

4 0

Answer:

Younger by nanoseconds

Explanation:

Not entirely sure :O

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Which is an important step in how an electric motor uses magnetic force to produce motion?

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Answer:B.As the north pole of the electromagnet nears the south pole of the permanent magnet, the current reverses and the poles of the magnets then repel.

Explanation:

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2 years ago

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A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme

Shkiper50 [21]

Answer: The density of this piece of jewelry is $8.90g/cm^3$

Explanation:

To calculate the density, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = $14.7cm^3$ $1cm^3=1ml$

Putting values in above equation, we get:

$\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3$

Thus density of this piece of jewelry is $8.90g/cm^3$

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2 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

(c) $\mu=9.71*10^{-24}A\cdot m^2$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

(c) Magnetic moment is given by:

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

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2 years ago

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. W

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Answer:

31

Explanation:

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3 years ago

A heavy piece of hanging sculpture is suspended by a 90 cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at it

kupik [55]

Answer: The mass of the sculpture is 11.8kg

Explanation:

Using the equation of fundamental frequency of a taut string.

f = (1/2L)*√(T/μ) .... (Eqn1)

Where

f= frequency in Hertz =80Hz

T = Tension in the string = Mg

M represent the mass of the substance (sculpture) =?

g= 9.8m/s^2

L= Length of the string=90cm=0.9m

μ= mass density = mass of string /Length of string

mass of string =5g=0.005kg

L=0.9m

μ=0.005/0.9 = 0.0056kg/m

Using (Eqn1)

80= 1/(2*0.9) √(T/0.0056)

144= √(T/0.0056)

Square both sides

20736= T/0.0056

T= 116.12N

Recall that T =Mg

116.12= M * 9.8

M=116.12/9.8

M= 11.8kg

Therefore the mass of the sculpture is 11.8kg

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2 years ago