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Hitman42 [59]
3 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]3 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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Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

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serg [7]

Answer:

a) v = 6.43 m/s

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For v relative to the car is we have vector sum of both velocities

v = √(14.44^2 + 6.43^2) = 15.8 m/s which is the velocity relative to car

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