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bearhunter [10]
4 years ago
12

Does anybody know about free fall for physics

Physics
2 answers:
melisa1 [442]4 years ago
8 0

Answer:

61.74 m/s

in short vf= 9.8m/s * 6.3 s

then u get 61.74

Explanation:

in Newtonian physics , free fall is any motion of a body where gravity is the only acceleration acting upon it . In the context of general relativity , where gravitation is reduced to a space - time curvature , a body in free fall has no force acting upon it.

thanks i hope it helped.

Licemer1 [7]4 years ago
3 0

Answer: 61.74 m/s

Explanation: Use the first equation for free falling bodies

a = vf - vi / t

Derive for vf:

vf = at + vi ( vi = 0 since it starts from rest)

Substitute the values

vf = 9.8 m/s² x 6.3 s

= 61.74 m/s

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Plzzzz urgent <br><br><br>solve this​
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\large{ \tt{☄ \: EXPLANATION}} :

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  • Notice that we're provided the time ( t ) in minutes. So , first thing we have to do is convert the minutes into seconds. It would be - Time ( t ) = 5 minutes = 5 × 60 sec = 300 sec [ 1 min = 60 sec ]

  • Here , We're provided - Initial velocity ( u ) = 0 , Final velocity ( v ) = 60 m / s , Time taken ( t ) = 300 seconds & We're asked to find out the acceleration ( a ) & distance covered by the jeep ( s ) .

\large{ \tt{♨ \:LET'S \: START}} :

  • Acceleration is defined as the rate of change of velocity. We know :

\large{ \boxed{ \tt{❁ \: ACCELERATION \: (a) =  \frac{FINAL \: VELOCITY(v) - INITIAL \: VELOCITY(u)}{TIME \: TAKEN \: ( \: t \: )}}}}

- Plug the values & then simplify !

\large{ \bf{↬a =  \frac{60 - 0}{300}  =  \frac{60}{300} =  \boxed{ \bold{ \bf{0.2 \: m {s}^{ - 2} }}} }}

  • The acceleration of the jeep is 0.2 m/s²

\large{ \tt{۵ \: AGAIN, \: USING\: SECOND \: EQUATION \: OF \: MOTION}} :

\boxed{ \large{ \bf{✾ \: s =  \frac{u + v}{2}  \times t}}}

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\large{ \bf{↦s =  \frac{0 + 60}{2}  \times 300 =  \boxed{ \bold{ \bf{9000 \: m}}}}}

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A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature
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<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

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The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

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q = heat absorbed or released

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T_1 = initial temperature of ice = -15.00°C

T_2 = initial temperature of water = 25.00°C

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