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3 years ago

Science- if you lived by the sea, what effect of the moon would you see? Describe the effect.

1 answer:

3 0

It would be waves because the moon creates a gravitational pull on the body of water. The closer or larger they are, the greater the pull. Each day there are 2 high tides and 2 low tides. The wind and the currents make waves and the gravitational pull makes the ocean bulge out towards the moon. Ocean levels change everyday as the sun, moon, and earth interact. There are 2 types of tides, spring tides and neap tides. Spring tides occur during a full moon and new moon, and neap tides occur during quarter moons.

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Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced fro

soldier1979 [14.2K]

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of copper wire,d=0.0113 in

Radius of copper wire= $r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in$

$\frac{1}{1000}=10^{-3}$

Density of copper= $\rho=8.96g/cm^3$

1 lb=454 g

3.25 lb= $3.25\times 454=1475.5 g$

Mass of Azurite= $1475.5 g$

Mass of copper= $\frac{55.1}{100}\times 1475.5=813 g$

Density= $\frac{Mass}{volume}$

Using the formula

$8.96=\frac{813}{volume\;of\;copper}$

Volume of copper wire= $\frac{813}{8.96}=90.7cm^3$

Radius of copper wire= $5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm$

1 in=2.54 cm

Volume of copper wire= $\pi r^2 h$

$\pi=3.14$

Using the formula

$90.7=3.14\times (14.35\times 10^{-3})^2\times h$

$h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}$

$h=140273 cm$

1 m=100 cm

$h=\frac{140273}{100}=1402.73 m$

Hence, the length of copper wire required=1402.73 m

7 0

3 years ago

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For maximum emf, $\sin\omega t=1$

So,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

So, the maximum emf generated in the loop is 5.32 V.

3 0

3 years ago

Every 6 seconds a pendulum completes 1 cycle. What are the period and frequency of this pendulum?

Kruka [31]

Multiply 6 x 60 to get the frequency.

8 0

3 years ago

Which of the following describes resistance force?

Harman [31]

I am pretty sure that<span> the following which describes resistance force is the fourth option from the scale represented above : </span>D .force applied by the machine to overcome resistance. I choose this due to the Newton's 3rd law, as the<span> force that shoul be overcome by a machine before it perform its usual work. Do hope you will find it helpful! Regards!</span>

4 0

3 years ago

Read 2 more answers

PLease answer this question or explain

scoray [572]

Answer:

NOAA Scienctist

Explanation:

8 0

3 years ago