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3 years ago

What is the special feature of the upper mantle?

1 answer:

mote1985 [20]3 years ago

3 0

The special feature of the upper mantle is the asthenosphere.It is located just below the lithosphere and is made up of rocks that is fluid and can move.The fluidity of these rocks powered the movement of the tectonic plates on the earths crust.Circular convection cycles in the hot,fluid upper mantle rock move the plates over the surface of the earth.

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Plane-polarized light is incident on a single polarizing disk, with the direction of E0 parallel to the direction of the transmi

Vladimir [108]

Answer: $65.9^{\circ}$

Explanation:

Given

Intensity must be reduced by a factor of 6

Intensity is given by $I=I_o\cos ^2\theta\\$

Substitute $I$ by $\frac{I_o}{6}$

$\Rightarrow \dfrac{I_o}{6}=I_o\cos ^2\theta\\\\\Rightarrow \cos^2\theta =\dfrac{1}{6}\\\\\Rightarrow \cos \theta=\dfrac{1}{\sqrt{6}}\\\\\Rightarrow \theta=65.9^{\circ}$

So, the disk must be rotated by an angle of $65.9^{\circ}$ .

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2 years ago

Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the

melisa1 [442]

Which behavior would best describe someone who has good communication skills with customers ? a) Following up with some customers b) Talking to customers more than listening to them c) Repeating back what the customer says d) Interrupting customers frequently

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3 years ago

Why Physical Education is Important in School?

blsea [12.9K]

Answer:

Physical education provides cognitive content and instruction designed to develop motor skills, knowledge, and behaviors for physical activity and physical fitness. Supporting schools to establish physical education daily can provide students with the ability and confidence to be physically active for a lifetime.

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2 years ago

Read 2 more answers

A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$

The CCD sensor is 35.34 mm from the lens

b) Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}$

$m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm$

The person appears 17.67 mm tall on the sensor

7 0

2 years ago

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal

Arturiano [62]

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² = 5880000 / 0.018

ω ² = 326666667

ω = 18074 rad/sec

7 0

2 years ago