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Ksenya-84 [330]
3 years ago
13

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

l as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×10²²kg and a radius of 1821 km. For this calculation, ignore any variation in gravity over the 500 km range of the debris. How high would this material go on earth if it were ejected with the same speed as on Io?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

H_2 = 91.55 km

Explanation:

Gravity on the surface of planet is given as

g = \frac{GM}{R^2}

as we know that

M = 8.93 \times 10^{22} kg

R = 1821 km

now gravity on the planet is

g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}

so we have

g = 1.8 m/s^2

now we know that

H_{max} = \frac{v^2}{2g}

so we will say

\frac{H_1}{H_2} = \frac{g_2}{g_1}

\frac{500}{H_2} = \frac{9.81}{1.8}

H_2 = 91.55 km

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How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?
soldi70 [24.7K]

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

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3 years ago
The mass of a golf ball is 45.9
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The equation for the de Broglie wavelength is: 
<span>λ = (h/mv) √[1-(v²/c²)], </span>
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<span>λ = h/mv. </span>
<span>Substituting, (remember to convert the mass to kg, since 1 J = 1 kg·m²/s²): </span>
<span>λ = (6.63x10^-34 J·s) / (0.0459 kg) (72.0 m/s) = 2.00x10^-34 m.</span>
8 0
3 years ago
A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0
3 years ago
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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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Why does rubbing a balloons on wool or hair make it attract other objects
KonstantinChe [14]

Answer:

its because static electricity my guy

Explanation:

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3 years ago
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