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Hitman42 [59]
2 years ago
15

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

heights of 70 km above the surface of this moon. If the value of g on Io is 2.0 m/s2 , estimate the speed with which the liquid sulfur left the volcano.
Physics
1 answer:
Y_Kistochka [10]2 years ago
7 0

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The speed with which the liquid sulfur left the volcano is 529.15 m/s

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A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of
kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0
2 years ago
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
SVETLANKA909090 [29]

Answer:

The  value is  v =  -0.04 \  m/s

Explanation:

From the question we are told that

   The  mass  of the block is  m  =  2.0 \ kg

   The  force constant  of the spring is  k  =  590 \ N/m

   The amplitude  is  A =  + 0.080

   The  time consider is  t =  0.10 \  s

Generally the angular velocity of this  block is mathematically represented as

      w =  \sqrt{\frac{k}{m} }

=>   w =  \sqrt{\frac{590}{2} }

=>   w = 17.18 \  rad/s

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         v  =  -A w sin (w* t )

=>       v  = -0.080 * 17.18 sin (17.18* 0.10 )

=>       v =  -0.04 \  m/s

7 0
2 years ago
As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre
diamong [38]

Answer:

USE SOCRACTIC IT WOULD REALLY HELP

4 0
2 years ago
Read 2 more answers
Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?
MakcuM [25]
The angle of refraction would be further less 
3 0
2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
2 years ago
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