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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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Potassium is a crucial element for the healthy operation of the human
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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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3 years ago
When force is applied to a breaker bar the torque can be calculated by multiplying the length of the lever by the?
Nimfa-mama [501]

When a force applied to a breaker bar the torque can be calculated by multiplying the<u> length of the lever</u> by the tangential component of force on the lever.

<h3>What is torque?</h3>

Torque is the <u>rotating equivalent</u> of force in physics and mechanics. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Torque is given by the formula :

                          α = r x F ( bold letters represent vector quantities)

The S.I. unit for torque is :  N - m ( Newton - meter)

<h3>How do we define 1 N-m of torque?</h3>

The newton-metre is a torque unit (also known as a moment) in the SI system. The torque produced by a one newton force applied <u>perpendicularly to the end of a one metre long</u> moment arm is known as a newton-metre.

To learn more about torque:

brainly.com/question/14970645

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