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solmaris [256]
2 years ago
13

You have just landed on planet x. you take out a 250-g ball, release it from rest from a height of 12.0 m, and measure that it t

akes 2.4 s to reach the ground. you can ignore any force on the ball from the atmosphere of the planet. how much does the 250-g ball weigh on the surface of planet x? 104.16667 incorrect: your answer is incorrect. n
Physics
2 answers:
nekit [7.7K]2 years ago
7 0

For the first step, since we know the time it takes to fall 12 meters, 
we can calculate the acceleration of gravity on Planet-X. 

           Falling distance = (1/2) (gravity) (time²)

                            (12 m)  =  (1/2) (gravity) (2.4 seconds)²

                                        =   (1/2 gravity)  (5.76 seconds²)

Divide each side by  (5.76 s²) :

                       1/2 gravity  =  (12 m) / (5.76 s²)

                             gravity  =  (24 m) / (5.76 s²)

                           Gravity  =  4.17 m/s²     (42.5% of Earth gravity)

Now that we know the acceleration  of gravity on Planet-X,
we can calculate the weight of the ball, (or of any mass).

           Weight of any mass = (mass) x (gravity)

For the ball,     Weight = (0.250 kg) x (4.17 m/s²)

                                    =   1.042 Newton

(The same 250-gram ball would weigh 2.45 newtons on Earth.)
DENIUS [597]2 years ago
5 0
I believe your answer is off by a couple of decimal points. The answer is approximately 1.041 N
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Answer and Explanation:

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2 years ago
Analysis of the relationship between the fuel economy​ (mpg) and engine size​ (liters) for 35 models of cars produces the regres
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Answer:

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From the information given in question we have

Regression equation is : model- mpg = 36.44 - 3.829\times engine\ size

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8 0
3 years ago
Could anyone help with this? :)
bonufazy [111]
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An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless
Elenna [48]

Answer:

Explanation:

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= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

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.5 x 10 v² = .5 x 500 x .75²

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b ) kinetic energy of mass at the bottom

= /2 m v²

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c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45  - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

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I think the correct answer is B.

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