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AlladinOne [14]
3 years ago
14

Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l

. a third particle is placed between particles 1 and 2. where on the x axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? express your answer in terms of l.
Physics
1 answer:
wlad13 [49]3 years ago
8 0

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 


x = L(sqrt2 - 1) </span>

 

= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

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Answer:

v=0.3381\ m.s^{-1} in the direction of motion of Jacob

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mass  of Jacob, m_j=45\ kg

velocity of Jacob, v_j=8.08\ m.s^{-1}

mass of Ethan, m_e=31\ kg

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Now using the conservation of linear momentum for the case:

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5 0
3 years ago
Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. this can be represented by the fun
Romashka [77]

The values of t are <u>4.643 second</u> for the function H(t)=-16t^2+73t+6

What is batter misses?

An out in baseball happens when the umpire declares a batter or baserunner out. A hitter or runner who is out is no longer able to score runs and must go back to the dugout until their subsequent turn at bat. The batting team's turn is over after three outs are recorded in a half-inning.

In order to signal an out, umpires typically make a fist with one hand and then flex that arm, either upward on pop flies or forward on regular plays at first base. To indicate a called strikeout, home plate umpires frequently use a "punch-out" action.When a batter is struck by a pitched ball without making a swing at it, it is referred to as a hit-by-pitch. He consequently gets first base.

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula H(t)=-16t^2+vt+s

H(t)=-16t^2+73t+6

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

-16t^2+73t+6=0

We solve the equation using quadratic formula x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Substituting the values a=-16, b= 73, c=6

t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643

Learn more about the batter misses with the help of the given link:

brainly.com/question/19475098

#SPJ4

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