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andrey2020 [161]
3 years ago
7

This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ___________

_. This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ____________. have a component along the direction of motion remain perpendicular to the direction of motion have a component against the direction of motion first have a component along the direction of motion; then against it; then along it; etc.
Physics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

have a component along the direction of motion that remains perpendicular to the direction of motion

Explanation:

In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.

circular movement

          F = m a

          a = v² / r

          F = m v²/R

where the force is perpendicular to the velocity, all the force is used to change the direction of the velocity

in linear motion

         F = m a

where the force is parallel to the acceleration of the body, the total force is used to change the modulus of the velocity

the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion

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magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
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Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

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3 years ago
What is the formula for velovity?
GarryVolchara [31]

Answer:

if it's velocity u talking of.....

Explanation:

then it's displacement/ time

4 0
2 years ago
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