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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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klio [65]

Answer:

25.48

Explanation:

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What effect do shiny metals have on radiant energy?
Andreyy89

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Metals conduct heat and reduce the kinetic energy within the components that need to remain cool

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Read 2 more answers
2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
10 months ago
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0
2 years ago
A train travels 77 kilometers in 1 hour, and the 66 kilometers in 1 hour. What is the average speed?
Kipish [7]

Average speed = (total distance covered) / (time to cover the distance)

Total distance = (77km + 66km) = 143 kilometers

Time to cover the distance = 2 hours

Average speed = (143 km) / (2 hours) =  71.5 km per hour
6 0
3 years ago
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