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3 years ago

The San Andreas fault is an example of which type of tectonic plate boundary?

2 answers:

4 0

The San Andreas fault is an example of Transform tectonic plate boundary.
<span>The San Andreas fault is a long narrow valley where the plates that make up the fault meet. It consists of the Pacific Plate and the North American Plate. These plates slide past each other when they move. It is also known as a strike-slip fault. As the plates bump and jam against each other as they slide past each other, earthquakes jar the regions around these plates.</span>

Luba_88 [7]3 years ago

3 0

The San Andreas fault is an example of Transform tectonic plate boundary.
<span>The San Andreas fault is a long narrow valley where the plates that make up the fault meet.</span>

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A light wave travels through air in equals 1.00 at an angle of 35 degrees what angle does it have when it passes from the air in

lyudmila [28]

Answer: Angle 59 degree

Explanation: Given that the

n1 = 1.0

n2 = 1.5

Øi = 35 degree

From Snell law, which says that

n1/n2 = sinØ1/ sinØ2

Substitute all the parameters into the formula

1/1.5 = sin 35/sinØ2

Cross multiply

Sin Ø2 = 1.5 sin35

SinØ2 = 1.5 × 0.573 = 0.860

Ø2 = sin^-1(0.860)

Ø2 = 59.36 degree

Ø2 = 59 degree ( approximately)

It has angle 59 degree when passing from air to glass

5 0

3 years ago

2. Which statement is correct about the Krebs cycle?

IRINA_888 [86]

The correct option is

a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.

Explanation:

The Krebs citric acid cycle happens within the mitochondrial matrix and generates a pool of energy (ATP, NADH, and FADH2) from the oxidization of pyruvate, the tip product of metabolism. Pyruvate is transported into the mitochondria and loses dioxide to make acetyl-CoA, a 2-carbon molecule.

5 0

3 years ago

Read 2 more answers

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, $mg < ks_s$

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

$P = mv$

$P = 0$

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

$F_{spring} > mg$

$Ks > mg$

6 0

3 years ago

What is the definition of the half-life of a radioactive isotope?

Illusion [34]

Answer:Half-life is the amount of time it takes for the initial mass of the isotope to decompose, by half, into other lighter atoms.

Explanation:Different radioactive isotopes have different half-lives. For example, the element technetium-99m has a half life of 6 hours. This means that is 100 kg of the element is left to decay, in 6 hours, 50kg of the mass will have changed into other elements/atoms. The half-life of uranium-238 is 4.5 billion years while that of polonium-216 is only 0.145 seconds.

5 0

3 years ago

Read 2 more answers

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago