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Ray Of Light [21]
3 years ago
7

The first active volcano observed outside the Earth was discovered in 1979 on Io, one of the moons of Jupiter. The volcano was o

bserved to be ejecting material to a height of about 3.00 E5 m. Given that the acceleration of gravity on Io is 1.80 m/s2, find the initial velocity of the ejected material.
Physics
1 answer:
meriva3 years ago
8 0

Answer:

1039.23 m/s

Explanation:

Given:

Height, H = 3 x 10⁵m

acceleration of gravity on lo, a = 1.80 m/s² toward moon's surface i.e -1.80m/s² with respect to the earth

Now, at its highest point the velocity i.e the final velocity 'v' = 0m/s.

Now using the Newton's equation of motion

v² - u² =2as

where,

v = final velocity

u = initial velocity

a = acceleration

s = distance

now,

substituting the values we get

0²-u² = 2 × (-1.80) × 3 x 10⁵

or

u=\sqrt{10.8\times 10^5}=1039.23m/s

Hence, the <u>initial velocity</u><u> with which the material is ejected is </u><u>1039.23 m/s</u>

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Describe three important ways we use the electromagnetic spectrum in our everyday lives.
OlgaM077 [116]

Answer:

We use X-rays to help the injured, Radiowaves to communicate or entertain, and Visible light to see.

5 0
3 years ago
Dolphins communicate using compression waves (longitudinal waves). Some of the sounds dolphins make are outside the range of hum
lana66690 [7]

Answer: send the message underwater because a more dense medium would make the sound travel faster.

Explanation:

Dolphins communicate using compression waves - longitudinal waves. Longitudinal waves requires a medium to travel. A longitudinal wave transfers energy by the vibration of medium particles in the direction of the wave motion. Compression are the regions where density of the medium is higher and rarefaction is a low density region.

A longitudinal wave travels faster in a denser medium. It has maximum speed in solid and minimum in gas. Thus, to transfer message quickly to dolphin B., dolphin A should send the message underwater and not in air. This is because water has higher density than air. Molecules collide more quickly in water than in air and it takes less time for signal to travel.

3 0
3 years ago
Three equal point charges, each with charge 2.00 μC , are placed at the vertices of an equilateral triangle whose sides are of l
oksano4ka [1.4K]

Answer:

Explanation:

Energy of system of charges

= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂

q₁ , q₂ and q₃ are charges and  r₁₂ , r₁₃ , r₃₂ are densities between them

9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )

= 9 x 10⁹  x 3 x 16 x 10⁻¹²

= 432 x 10⁻³

= .432 J .

4 0
3 years ago
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
2 years ago
An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang
kirza4 [7]

Answer:

a=500m/s^2

Explanation:

We need only to apply the definition of acceleration, which is:

a=\frac{v_f-v_i}{t_f-t_i}

In our case the final velocity is v_f=10m/s, the initial velocity is v_i=0m/s since it departs from rest, the final time is t_f=0.02s and the initial time we are considering is t_i=0s

So for our values we have:

a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2

3 0
3 years ago
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