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1 year ago

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Seafloor spreading is a process that occurs along _____1. At _2________ boundary, two plates mov

e away from each other resulting to spreading of seafloor, thereby creating new ___________3____________ crust. Near the mid-oceanic ridge _________4__________ rocks are found and far from it are older.

1 answer:

sergejj [24]1 year ago

5 0

Answer:

where is the answer choices

Explanation:

You might be interested in

Is a mirrors surface transparent translucent or opaque how do you know.

Eva8 [605]

There are actually two different kinds of mirrors, and the answer is different
for each one.

-- Plain old everyday hand mirror, vanity mirror, bathroom mirror, makeup
mirror, etc.

Opaque, reflecting silver coating is on the back of the glass.
Light from your tongue or your teeth flows to the front surface of the glass,
through the glass, out of the back surface of the glass, bounces off of the silver
coating on the back, reverses its direction, enters the back surface of the glass,
comes back through the glass again, leaves the front of the glass, goes into your
eyes, and you can see your teeth or your tongue.

Both surfaces of the glass, as well as the glass in between the surfaces, are
transparent. The silver coating on the back is opaque. I know that, because
when I look at the back of a mirror, I can't see any light coming through it.
The coating on the back is also reflective ... a big part of the reason why
a mirror works.

-- Expensive mirrors used by astronomers and eye-doctors.
Known as "first surface" mirrors.

Opaque, reflecting silver coating is on the <em>front</em> of the glass.
Light from your tongue or your teeth flows toward the front surface of the glass,
but never actually gets there. It bounces off of the silver coating on the front of
the glass, reverses its direction, goes into your eyes, and you can see your teeth
or your tongue.

The glass is transparent, but that doesn't matter, because the light never reaches
the glass. It only goes as far as the opaque silver coating on the front, and is
reflected from there.

4 0

2 years ago

Can someone give me the units

Anettt [7]

Answer:

The unit for power is the Watt

8 0

2 years ago

10. when creating a cardiovascular fitness routine which program would be appropriate?

sasho [114]

Answer:

10. B

13. It's actually 30 seconds but 20 comes closest so B

14. B, any type of stretching is good before you perform any type of physical activity because it loosens up the muscles

15. D, all of the above

Explanation:

I have done many sports in my life and am still very active, I grew to know most of these things through the years.

5 0

2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

2 years ago

A4 kg bowling ball begins rolling down a at bowling alloy at 6 m/s . When it strikes the pins, it is estimated to be moving at 5

Paul [167]

Answer:

Energy lost due to friction is 22 J

Explanation:

Mass of the ball m = 4 kg

Initially velocity of ball v = 6 m/sec

So kinetic energy of the ball $KE=\frac{1}{2}mv^2$

$KE=\frac{1}{2}\times 4\times 6^2=72J$

Now due to friction velocity decreases to 5 m/sec

Kinetic energy become

$KE=\frac{1}{2}\times 4\times 5^2=50J$

Therefore energy lost due to friction = 72 -50 = 22 J

8 0

2 years ago