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3 years ago

Please sb help me fr

Physics

2 answers:

marissa [1.9K]3 years ago

7 0

me lo traduses plis no se ingles solo escribe lo que dice en la guia

DIA [1.3K]3 years ago

5 0

Answer:

I think it might be box 1. Hope this is right have a good day

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At depths of greater than 60 m, ice moves by plastic deformation. What causes this transition from brittle to ductile behavior?

mafiozo [28]

Answer:

Increase in the intensity of pressure allows very less deformation

Explanation:

For the water the variation of pressure at any depth is directly proportional to the depth of the water above that level.

Thus, at depth greater than 60 m the intensity of pressure on ice is very high and therefore very little deformation is possible.

Hence, the transition from brittle to ductile behavior is observed.

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4 years ago

Which best describes the net force on a pair of balanced forces?

Butoxors [25]

The net force of a pair of balanced forces is zero

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3 years ago

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An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the

elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

$F_c = F_g$

$\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}$

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

$r \rightarrow R+h \Rightarrow$ Radius of earth and orbit

Re-arrange to find the velocity

$\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}$

$\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}$

$v^2_{orbit}=\frac{GM}{r}$

$v_{orbit} = \sqrt{\frac{GM}{r}}$

$v_{orbit} = \sqrt{\frac{GM}{R+h}}$

Replacing with our values we have

$v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}$

$v_{orbit} = 7676m/s$

From the cinematic equations of motion we have to

$t = \frac{d}{2v_{orbit}} \rightarrow$ Remember that the speed is double for the counter-direction of the trajectories.

Replacing

$t = \frac{29000m}{7676m/s}$

$t = 3.778s$

Therefore the time required is 3.778s

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3 years ago

Under what atmospheric condition is fog most often formed in the san joaquin valley?

GalinKa [24]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which <span>fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.</span>

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4 years ago

Which energy transformation pops the corn kernels

Andrews [41]

Answer:

heat?

Explanation:

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3 years ago