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3 years ago

What is a tsunami? How is it formed?

2 answers:

4 0

A tsunami is a series of waves generated in an ocean or other body of water by a disturbance such as an earthquake, landslide, volcanic eruption, or meteorite impact. ... Undersea earthquakes, which typically occur at boundaries between Earth's tectonic plates, cause the water above to be moved up or down

PtichkaEL [24]3 years ago

4 0

Answers:

1. A tsunami, also known as a seaquake is a seism whose epicenter is located at the depths of the sea or ocean (or any a mass of water) producing the agitation of the waters and forming large waves.

2. How is it formed?

After the seism at the depth of the mass of water begins, and if the seism has a vertical component, the movement causes a displacement of the sea bottom and, consequently, it causes the formation giant waves as the water is displaced because of gravity, trying to return to a stable position.

Now, when the tsunami approaches the coast its speed decreases, because the depth decreases, in other words, the water accumulates when braking, increasing the height of the wave.

And, according to the conservation of energy principle, if the speed of the wave decreases, its height increases.

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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

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3 years ago

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?

grandymaker [24]

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

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3 years ago

If the tension of the cable is 25.0 N what is the mass of the ball

marysya [2.9K]

Answer:576

Explanation:

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3 years ago

When Dr. Montero was observing the endoplasmic reticulum of a cell using an electron microscope, she noticed that it was covered

Vesnalui [34]

The answer is Convoluted endoplasmic reticulum

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3 years ago

An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of

slega [8]

Answer:

$T_{1}=94.9^{o}C$

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

$T_{1}=T_{2}+\frac{q}{kS}$

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

$S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\$

Substitute the given values

$S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m$

The temperature of heater is then:

$T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C$

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

$T_{1}=94.9^{o}C$

5 0

3 years ago