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chubhunter [2.5K]
3 years ago
10

What is a relatively soft layer of upper mantle called?

Physics
1 answer:
brilliants [131]3 years ago
3 0
The answer is asthenosphere

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A person makes a cup of coffee by first placing a 200 W electric immersion heater in 0.32 kg
alexdok [17]

Answer:

Answer is D. 8.04 x 10^4 J

Explanation:

1. D

2. A

3. D

4. B

5. C

6. B

7. D

8. C

9. B

10. D

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8 0
3 years ago
As the average kinetic energy of a substance increases, the tempature of the sample
taurus [48]
I have no clue. Sorry bro
6 0
2 years ago
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A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp
NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

f=12GHz=12\cdot 10^9 Hz is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

So the wavelength of the beam is

\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

and so, the diameter is

d=2y = 750 m

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

r=\frac{750 m}{2}=375 m

So the area is

A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2

And since the power is

P=100 kW = 1\cdot 10^5 W

The average intensity is

I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2

4 0
3 years ago
A(n) _____ is an organic compound that changes color in acid or bases.
Romashka-Z-Leto [24]
The answer is indicator.
6 0
3 years ago
three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
2 years ago
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