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3 years ago

Volcanologists and seismologist are specialzed in what

Physics

2 answers:

antiseptic1488 [7]3 years ago

5 0

The answer is:

They are specialized in geology and geophysics

1) Volcanologists are scientific researchers who study the properties of volcanoes and develop the science of predicting volcanic eruptions accurately.

- Volcanoes are the results of fissures in the earth's crust, from which magma, gases, rocks, and ashes can erupt.

-Volcanic eruptions and their impact on the atmosphere are investigated by specialist geologists known as volcanologists.

A) Volcanologists are scientists who use a variety of sophisticated equipment to measure and analyze volcanic activity, lava, rock, ashes and gases as well as earthquakes caused by eruptions.

B) Also, they measure the magnitude and intensity of explosions, placing them on a scale from 1 to 8 [volcanic explosivity index].

C)Volcanologists monitor both active and dormant volcanoes to try to predict eruptions and minimize adverse effects on people and their environment.

D)Also, they spend a lot of time collecting and analyzing samples of debris, rocks and other products from extinct volcanoes to gain a better understanding of the processes behind the volcanic activity and the impact former eruptions had in shaping the climate.

2) Seismologists are Earth scientists, specialized in geophysics, who study the genesis and the propagation of seismic waves in geological materials.

-Their research aims at interpreting the geological composition and structures of the Earth. In the case of earthquakes, seismologists evaluate the potential dangers and seek to minimize their impact through the improvement of construction standards.

AysviL [449]3 years ago

4 0

Volcanologists and seismologists are specialized in geology

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An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

maw [93]

Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

4 0

3 years ago

Coil the wire around the nail. Connect one end of the wire to the positive battery terminal. Connect the other end of the wire t

Andrej [43]

Static i think
not sure
hope it helps

5 0

3 years ago

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther

jeka94

Answer:

T = 188.5 s, correct is C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

angular and linear velocity are related

v = r w

w = v / r

m r v₀ = I₀ v / r + m r v

m r v₀ = (I₀ / r + mr) v

v = $\frac{m}{\frac{I_o}{r} +mr} \ r v_o$

let's calculate

v = $\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4$

v = $\frac{0.020}{2.345} \ 2.4$

v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

v = x / T

T = x / v

the distance of a circle with radius r = 0.6 m

x = 2π r

we substitute

T = 2π r / v

let's calculate

T = 2π 0.6/0.02

T = 188.5 s

reduce

t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is C

6 0

3 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

8 0

3 years ago

Read 2 more answers

What scientific principle causes the different layers of the earth

Vedmedyk [2.9K]

Answer: The stratification of earth into its geologic layers was brought about by th e formation of the earth iron core.

Explanation:

5 0

3 years ago