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3 years ago

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What is the acceleration of a baseball that has a final velocity of 15.0 m/s when it crosses the plate 2.0 seconds after leaving

the pitchers hand? The initial velocity is 0.0 m/s and the baseball is traveling forward.

1 answer:

Leokris [45]3 years ago

3 0

I think it’s A pretty sure...

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If a net force of 250 N causes an object to accelerate at 20m/s^2 what must its mass be?

aleksandr82 [10.1K]

The answer is 12.5 kg because 250N / 20m/s^2

I hope that helped

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Give me 1 example of complete inelastic collision

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2 years ago

I NEED THIS ASAP!!! 20 points!

suter [353]

Gravitational potential energy=mass*gravitational acceleration*heightKinetic energy = 0.5*mass*velocity²Thus:K.E0.5*1*x²=12.5x²=12.5/(0.5*1)x=√12.5/(0.5*1)x=5
GPEmass*gravitational acceleration*height1*9.81*h=98h=98/(9.81*1)h= 9.98 J approximately, rounded 10meters

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3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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2 years ago