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olga_2 [115]
3 years ago
6

The first active volcano observed outside the earth was discovered in 1979 on io, one of the moons of jupiter. suppose a volcano

on another moon is observed to be ejecting material to a height of about 1.52 ✕ 105 m. given that the acceleration of gravity on this moon is 1.91 m/s2 find the initial velocity of the ejected material.
Physics
1 answer:
inna [77]3 years ago
4 0

Gravitation due to the planet's mass is always against the ejected particle velocity. This will result in deceleration of the ejected particle and eventually bring the velocity to zero at the topmost point of the trajectory before pulling it back to the ground.

By the kinematic equation

v^2 - u^2 = 2.a.s

where s is the distance travelled = 1.52 x 10^5 m

a is the acceleration due to gravity = -1.91 m/s^2

v and u are initial and final velocities, so v = 0.

Therefore

u = \sqrt{s.a.s}

u = 761.99 m/s

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The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed?(a) 16(b
maw [93]

The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

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The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

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4 0
1 year ago
A giant log float on a lake while a tiny grain of sand sinks to the bathroom. Explain why a heavy object like the log floats whi
ella [17]

Answer:

Because the log is hollow or is less dense

Explanation:

If a log has little density, it will float. While sand is dense and completely solid, so it will sink.

7 0
3 years ago
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The coefficient of friction between a rider and the merry go round is 0.45 and the person is measured to be traveling at 20.0 RP
svlad2 [7]

The person must stand at a radius of 0.99 m

Explanation:

In order for the person to stand on the merry go round, the force of friction acting on the person must provide the centripetal force necessary to keep the person in uniform circular motion.

Therefore, we can write:

\mu mg = m\omega^2 r

where:

- the term on the left is the force of friction, and the term on the right is the centripetal force

\mu = 0.45 is the coefficient of friction

m is the mass of the person

g=9.8 m/s^2 is the acceleration of gravity

\omega = 20 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=2.09 rad/s is the angular velocity

r is the radius of the circular path

Solving the equation for r, we find the radius at which the person must be standing:

r=\frac{\omega^2}{\mu g}=\frac{(2.09)^2}{(0.45)(9.8)}=0.99 m

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8 0
3 years ago
On a beautiful night in Washington D.C., you see a mirror image of the Washington Monument and surrounding scenery. What causes
Alchen [17]
Your answer would be B, Reflected light! hope this helps
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4 years ago
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The position of an electron is measured within an uncertainty of 0.100 nm. What will be its minimum position uncertainty 2.00 s
Tanya [424]

Answer:

Minimum uncertainty in position is \Delta x= 1157808.48\ m

Explanation:

It is given that,

Uncertainty in the position of an electron, \Delta x=0.1\ nm=0.1\times 10^{-9}\ m

According to uncertainty principle,

\Delta x.\Delta p\geq \dfrac{h}{4\pi}

\Delta x.m\Delta v\geq \dfrac{h}{4\pi}

\Delta v\geq \dfrac{h}{4\pi \times \Delta x\times m}

\Delta v\geq \dfrac{6.62\times 10^{-34}\ J-s}{4\pi \times 0.1\times 10^{-9}\ m\times 9.1\times 10^{-31}\ kg}

\Delta v\geq 578904.24\ m/s

Let \Delta x is the uncertainty in position after 2 seconds such that,

\Delta x=\Delta v\times t

\Delta x=578904.24\ m/s\times 2\ s

\Delta x= 1157808.48\ m

or

\Delta x= 1.15\times 10^6\ m

Hence, this is the required solution.

7 0
3 years ago
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