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antiseptic1488 [7]
3 years ago
15

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:
alisha [4.7K]3 years ago
6 0

Answer:

91.54 km

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Io = 1821000 m

m = Mass of the Io =  8.93 × 10²² kg

H_i = 500000 m

g = Acceleration due to gravity on Earth = 9.81 m/s²

Acceleration due to gravity on Io

g_i=\frac{GM}{r^2}\\\Rightarrow g_i=\frac{6.67\times 10^{-11}\times 8.93\times 10^{22}}{1821000^2}\\\Rightarrow g_i=1.796\ m/s^2

H_e=\frac{v^2}{2g}

H_i=\frac{v^2}{2g_l}

Dividing the above equations we get

\frac{H_e}{H_i}=\frac{\frac{v^2}{2g}}{\frac{v^2}{2g_i}}\\\Rightarrow \frac{H_e}{H_i}=\frac{g_i}{g}\\\Rightarrow \frac{H_e}{H_i}=\frac{1.796}{9.81}\\\Rightarrow H_e=H_i\times \frac{1.796}{9.81}\\\Rightarrow H_e=500000\times \frac{1.796}{9.81}\\\Rightarrow H_e=91539.245\ m=91.54\ km

The material would go 91.54 km on earth if it were ejected with the same speed as on Io.

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What units are used to measure and communicate the amount of a force?
jeyben [28]

Answer:

The SI unit of force is the newton, symbol N. The base units relevant to force are: The metre, unit of length — symbol m. The kilogram, unit of mass — symbol kg.

6 0
3 years ago
The Pacific Giant Kelp grows at a rate of 18 in/day. How many centimeters per hour is this?
Airida [17]

For this case we must do a conversion. By definition we have to:

1 inch equals 2.54 centimeters

1 day equals 24 hours.

So, we have:18 \frac {in} {day} * \frac {1} {24} \frac {day} {h} * \frac {2.54} {1} \frac {cm} {in} =

Canceling units we have:

\frac {18 * 1 * 2.54} {24 * 1} \frac {cm} {h} =\\\frac {45.72} {24}\frac {cm} {h} =

1,905 \frac {cm} {h}

Answer:

1,905 \frac {cm} {h}

5 0
3 years ago
The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-
WINSTONCH [101]

Answer: 1.212(10)^{-10} m

Explanation:

The de Broglie wavelength \lambda is given by the following formula:

\lambda=\frac{h}{p} (1)

Where:

h=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

p is the momentum of the atom, which is given by:

p=m_{e}v (2)

Where:

m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg is the mass of the electron

v=6(10)^{6}m/s is the velocity of the electron

This means equation (2) can be written as:

p=(9.11(10)^{-31}kg)(6(10)^{6}m/s) (3)

Substituting (3) in (1):

\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)} (4)

Now, we only have to find \lambda:

\lambda=1.2122(10)^{-10} m>>> This is the de Broglie wavelength of the electron

8 0
3 years ago
Suppose a piano tuner hears 2 beats per second when listening to the combined sound from her tuning fork and the piano note bein
maksim [4K]

Answer:

further tightening is required.

Explanation:

The beat created / sec = difference of frequencies

Initial beat heard = 2

so difference of frequencies = 2

after tightening beat heard  = 1

difference  of frequencies decreases because frequency of tuning fork was higher than piano sound.

on further tightening  difference decreases because tightening increases the frequency  of piano hence   further  tightening is required for resonance.

6 0
2 years ago
A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling
ivann1987 [24]

Answer:

42.99°

Explanation:

F_h = Kinetic friction force

F_{\theta} = Pulling force at angle \theta

N_h = Weight of the box = 150 N

Kinetic friction force

F_h=\muN_h

Pulling force at angle \theta

F_{\theta}=\muN_{\theta}

N = Pulling force

According to question

\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N

Applying Newton's second law in the vertical direction we get

N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}

The angle is 42.99°

8 0
3 years ago
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